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		<title>Funci&#243;n de onda de una part&#237;cula cu&#225;ntica como combinaci&#243;n lineal de los estados estacionarios (8466)</title>
		<link>https://ejercicios-fyq.com/Funcion-de-onda-de-una-particula-cuantica-como-combinacion-lineal-de-los</link>
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		<dc:date>2025-05-29T11:34:42Z</dc:date>
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&lt;p&gt;Considera una part&#237;cula cu&#225;ntica de masa &#171;m&#187; confinada en un pozo de potencial unidimensional infinito en el intervalo . En el instante t = 0, la funci&#243;n de onda de la part&#237;cula viene dada por: &lt;br class='autobr' /&gt; &lt;br class='autobr' /&gt;
a) Normaliza la funci&#243;n de onda inicial y verifica que ya est&#225; normalizada. &lt;br class='autobr' /&gt;
b) Expresa como una combinaci&#243;n lineal de los estados estacionarios del pozo infinito. &lt;br class='autobr' /&gt;
c) Determina la funci&#243;n de onda en un tiempo t &gt; 0. &lt;br class='autobr' /&gt;
d) Calcula la probabilidad de que, al medir la energ&#237;a, se obtenga el (&#8230;)&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Fisica-cuantica-288" rel="directory"&gt;F&#237;sica cu&#225;ntica&lt;/a&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Considera una part&#237;cula cu&#225;ntica de masa &#171;m&#187; confinada en un pozo de potencial unidimensional infinito en el intervalo &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L93xH18/db921407c58886ead57ef43e7f1adfac-b6685.png?1748518547' style='vertical-align:middle;' width='93' height='18' alt=&#034;0 \leq x \leq L&#034; title=&#034;0 \leq x \leq L&#034; /&gt;. En el instante t = 0, la funci&#243;n de onda de la part&#237;cula viene dada por:&lt;/p&gt;
&lt;p&gt;
&lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L224xH52/3f49bbefa9b2e07427d05c16ef3800da-6a053.png?1748518547' style='vertical-align:middle;' width='224' height='52' alt=&#034;\Psi(x, 0) = \sqrt{\frac{30}{L^5}} \, x (L - x)&#034; title=&#034;\Psi(x, 0) = \sqrt{\frac{30}{L^5}} \, x (L - x)&#034; /&gt;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;a) Normaliza la funci&#243;n de onda inicial y verifica que ya est&#225; normalizada.&lt;/p&gt;
&lt;p&gt;b) Expresa &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L61xH23/57b582839fc8384f7195f9066219374a-da523.png?1748518547' style='vertical-align:middle;' width='61' height='23' alt=&#034;\Psi(x, 0)&#034; title=&#034;\Psi(x, 0)&#034; /&gt; como una combinaci&#243;n lineal de los estados estacionarios &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L50xH23/0384894ad2cad3e86d932dbdb7202511-8059e.png?1748518547' style='vertical-align:middle;' width='50' height='23' alt=&#034;\psi_n(x)&#034; title=&#034;\psi_n(x)&#034; /&gt; del pozo infinito.&lt;/p&gt;
&lt;p&gt;c) Determina la funci&#243;n de onda &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L59xH23/7958d280ff667bd2eb74e389b9e8bb1d-16dec.png?1748518547' style='vertical-align:middle;' width='59' height='23' alt=&#034;\Psi(x, t)&#034; title=&#034;\Psi(x, t)&#034; /&gt; en un tiempo t &gt; 0.&lt;/p&gt;
&lt;p&gt;d) Calcula la probabilidad de que, al medir la energ&#237;a, se obtenga el valor correspondiente al primer estado excitado (n = 2).&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) La funci&#243;n de onda inicial estar&#225; normalizada cuando cumpla la condici&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/74d83981dcbb5ac46342991fae33c0e3.png' style=&#034;vertical-align:middle;&#034; width=&#034;212&#034; height=&#034;54&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\int_{0}^{L} |\Psi(x, 0)|^2\ dx = 1}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\int_{0}^{L} |\Psi(x, 0)|^2\ dx = 1}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Escribes la ecuaci&#243;n sustiyendo la funci&#243;n de onda: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d380354b0642b7832863c214b2e394ba.png' style=&#034;vertical-align:middle;&#034; width=&#034;484&#034; height=&#034;69&#034; alt=&#034;\int_{0}^{L} \left( \sqrt{\frac{30}{L^5}} \, x (L - x) \right)^2 dx = \color[RGB]{2,112,20}{\bm{\frac{30}{L^5} \int_{0}^{L} x^2 (L - x)^2\ dx}}&#034; title=&#034;\int_{0}^{L} \left( \sqrt{\frac{30}{L^5}} \, x (L - x) \right)^2 dx = \color[RGB]{2,112,20}{\bm{\frac{30}{L^5} \int_{0}^{L} x^2 (L - x)^2\ dx}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si desarrollas el cuadrado y divides en tres integrales: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/052bace1481ebdfd7b966aa53d779630.png' style=&#034;vertical-align:middle;&#034; width=&#034;447&#034; height=&#034;54&#034; alt=&#034;\frac{30}{L^5} \left[ L^2 \int_{0}^{L} x^2 \, dx - 2L \int_{0}^{L} x^3 \, dx + \int_{0}^{L} x^4 \, dx \right] = 1&#034; title=&#034;\frac{30}{L^5} \left[ L^2 \int_{0}^{L} x^2 \, dx - 2L \int_{0}^{L} x^3 \, dx + \int_{0}^{L} x^4 \, dx \right] = 1&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Las integrales son inmediatas y las calculas entre los l&#237;mites de integraci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/9fbd1e95e7c488b711dee201f9aa0601.png' style=&#034;vertical-align:middle;&#034; width=&#034;512&#034; height=&#034;55&#034; alt=&#034;\frac{30}{L^5} \left( L^2\cdot \frac{L^3}{3} - 2L\cdot \frac{L^4}{4} + \frac{L^5}{5} \right) = \color[RGB]{0,112,192}{\bm{\frac{30}{L^5} \left( \frac{L^5}{3} - \frac{L^5}{2} + \frac{L^5}{5} \right)}}&#034; title=&#034;\frac{30}{L^5} \left( L^2\cdot \frac{L^3}{3} - 2L\cdot \frac{L^4}{4} + \frac{L^5}{5} \right) = \color[RGB]{0,112,192}{\bm{\frac{30}{L^5} \left( \frac{L^5}{3} - \frac{L^5}{2} + \frac{L^5}{5} \right)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si sacas factor com&#250;n &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/bee16925f81c32af7649b0798ec0e7ae.png' style=&#034;vertical-align:middle;&#034; width=&#034;20&#034; height=&#034;19&#034; alt=&#034;L^5&#034; title=&#034;L^5&#034; /&gt; y simplificas: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/8237f800da8c3f9fd5b1f393f9101993.png' style=&#034;vertical-align:middle;&#034; width=&#034;562&#034; height=&#034;54&#034; alt=&#034;\frac{30\cdot \cancel{L^5}}{\cancel{L^5}} \left( \frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right) = 30 \left( \frac{10}{30} - \frac{15}{30} + \frac{6}{30} \right) = \fbox{\color[RGB]{192,0,0}{\bm{30 \left( \frac{1}{30} \right) = 1}}}&#034; title=&#034;\frac{30\cdot \cancel{L^5}}{\cancel{L^5}} \left( \frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right) = 30 \left( \frac{10}{30} - \frac{15}{30} + \frac{6}{30} \right) = \fbox{\color[RGB]{192,0,0}{\bm{30 \left( \frac{1}{30} \right) = 1}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Como puedes ver, la funci&#243;n de onda est&#225; normalizada. &lt;br/&gt; &lt;br/&gt; b) La ecuaci&#243;n de los estados estacionarios de un pozo infinito es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6603ae6c6e1bbe7601218ca4e625ebff.png' style=&#034;vertical-align:middle;&#034; width=&#034;271&#034; height=&#034;56&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\psi_n(x) = \sqrt{\frac{2}{L}}\cdot sen \left( \frac{n \pi x}{L} \right) }}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\psi_n(x) = \sqrt{\frac{2}{L}}\cdot sen \left( \frac{n \pi x}{L} \right) }}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Tienes que expresar la funci&#243;n de onda como combinaci&#243;n lineal de los estados estacionarios: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/62bfdf01cdba5772ee24b568fd94c83a.png' style=&#034;vertical-align:middle;&#034; width=&#034;241&#034; height=&#034;60&#034; alt=&#034;\color[RGB]{0,112,192}{\bm{\Psi(x, 0) = \sum_{n=1}^{\infty} c_n\cdot \psi_n(x)}}&#034; title=&#034;\color[RGB]{0,112,192}{\bm{\Psi(x, 0) = \sum_{n=1}^{\infty} c_n\cdot \psi_n(x)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Los coeficientes &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6f58730f154756d9dc7efb13fc938933.png' style=&#034;vertical-align:middle;&#034; width=&#034;19&#034; height=&#034;15&#034; alt=&#034;c_n&#034; title=&#034;c_n&#034; /&gt; los calculas de esta manera: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/7019abec8dc6385fb18ff9ca80096a5b.png' style=&#034;vertical-align:middle;&#034; width=&#034;676&#034; height=&#034;56&#034; alt=&#034;c_n = \int_{0}^{L} \psi_n^*(x)\cdot \Psi(x, 0)\ dx = \sqrt{\frac{2}{L}}\cdot \sqrt{\frac{30}{L^5}} \int_{0}^{L} x\cdot (L - x)\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx&#034; title=&#034;c_n = \int_{0}^{L} \psi_n^*(x)\cdot \Psi(x, 0)\ dx = \sqrt{\frac{2}{L}}\cdot \sqrt{\frac{30}{L^5}} \int_{0}^{L} x\cdot (L - x)\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si operas con las constantes que est&#225; fuera del integrando tienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/7473043a08b243cad69cab32cfbd79df.png' style=&#034;vertical-align:middle;&#034; width=&#034;387&#034; height=&#034;54&#034; alt=&#034;c_n = \frac{2 \sqrt{15}}{L^3} \int_{0}^{L} (Lx - x^2)\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx&#034; title=&#034;c_n = \frac{2 \sqrt{15}}{L^3} \int_{0}^{L} (Lx - x^2)\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La resoluci&#243;n de la integral la haces en dos partes: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0bbb93f42f526b017b79dd841a02231f.png' style=&#034;vertical-align:middle;&#034; width=&#034;554&#034; height=&#034;105&#034; alt=&#034;\left L\ \int_{0}^{L} x\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx = \dfrac{L^2 (-1)^{n+1}}{n \pi} \atop \int_{0}^{L} x^2\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx = \dfrac{2L^3 (-1)^{n+1}}{n \pi} - \dfrac{L^3 (2 - n^2 \pi^2 (-1)^n)}{n^3 \pi^3} \right \}&#034; title=&#034;\left L\ \int_{0}^{L} x\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx = \dfrac{L^2 (-1)^{n+1}}{n \pi} \atop \int_{0}^{L} x^2\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx = \dfrac{2L^3 (-1)^{n+1}}{n \pi} - \dfrac{L^3 (2 - n^2 \pi^2 (-1)^n)}{n^3 \pi^3} \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Combinas los t&#233;rminos anteriores y tienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a0e3c5da74602fd6a2a34982021711ad.png' style=&#034;vertical-align:middle;&#034; width=&#034;560&#034; height=&#034;54&#034; alt=&#034;c_n = \frac{2 \sqrt{15}}{\cancel{L^3}}\cdot \cancel{L^3} \left( \frac{(-1)^{n+1}}{n \pi} - \frac{2\cdot (-1)^{n+1}}{n \pi} + \frac{(2 - n^2 \pi^2 (-1)^n)}{n^3 \pi^3} \right)&#034; title=&#034;c_n = \frac{2 \sqrt{15}}{\cancel{L^3}}\cdot \cancel{L^3} \left( \frac{(-1)^{n+1}}{n \pi} - \frac{2\cdot (-1)^{n+1}}{n \pi} + \frac{(2 - n^2 \pi^2 (-1)^n)}{n^3 \pi^3} \right)&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si tienes en cuenta que para valores pares de &#171;n&#187; los t&#233;rminos se cancelan y simplificas, obtienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e6d8d177177f3de76378443068737276.png' style=&#034;vertical-align:middle;&#034; width=&#034;237&#034; height=&#034;53&#034; alt=&#034;\color[RGB]{0,112,192}{\bm{c_n = \frac{4 \sqrt{15}}{n^3 \pi^3} \left( 1 - (-1)^n \right)}}&#034; title=&#034;\color[RGB]{0,112,192}{\bm{c_n = \frac{4 \sqrt{15}}{n^3 \pi^3} \left( 1 - (-1)^n \right)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Por lo tanto, para valores pares de &#171;n&#187; los coeficientes son nulos y para valores impares de &#171;n&#187; obtienes: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/04c2a1b221d1d61b754b6118114834ff.png' style=&#034;vertical-align:middle;&#034; width=&#034;112&#034; height=&#034;44&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{c_n = \frac{8 \sqrt{15}}{n^3 \pi^3}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{c_n = \frac{8 \sqrt{15}}{n^3 \pi^3}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; c) La funci&#243;n de onda en funci&#243;n del tiempo, escrita como combinaci&#243;n lineal de los estados estacionarios, es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/03b030ad0729d2fb05ce484362f06f3e.png' style=&#034;vertical-align:middle;&#034; width=&#034;492&#034; height=&#034;64&#034; alt=&#034;\color[RGB]{192,0,0}{\bm{\Psi(x, t) = \sum_{n=1, 3, 5, \dots} \frac{8\sqrt{15}}{n^3\pi^3} \sqrt{\frac{2}{L}}\cdot sen \left( \frac{n \pi x}{L} \right) e^{\frac{-i E_n t}{\hbar}}}}&#034; title=&#034;\color[RGB]{192,0,0}{\bm{\Psi(x, t) = \sum_{n=1, 3, 5, \dots} \frac{8\sqrt{15}}{n^3\pi^3} \sqrt{\frac{2}{L}}\cdot sen \left( \frac{n \pi x}{L} \right) e^{\frac{-i E_n t}{\hbar}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; donde el t&#233;rmino &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/f68207972fe0c39be7798431a8afcc29.png' style=&#034;vertical-align:middle;&#034; width=&#034;24&#034; height=&#034;20&#034; alt=&#034;E_n&#034; title=&#034;E_n&#034; /&gt; es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/f2f52bd80229fffb3461b7c22699af1c.png' style=&#034;vertical-align:middle;&#034; width=&#034;133&#034; height=&#034;51&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; d) Como &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d42957487f34bea1601b6333cada1826.png' style=&#034;vertical-align:middle;&#034; width=&#034;58&#034; height=&#034;20&#034; alt=&#034;c_n = 0&#034; title=&#034;c_n = 0&#034; /&gt; para cualquier valor par de &#171;n&#187;, la probabilidad de encontrar &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/8e3b512c2f053602a180ee612fd581a6.png' style=&#034;vertical-align:middle;&#034; width=&#034;18&#034; height=&#034;15&#034; alt=&#034;E_2&#034; title=&#034;E_2&#034; /&gt; es nula, es decir: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/647b31c714d943188eaf53f7bde46f47.png' style=&#034;vertical-align:middle;&#034; width=&#034;196&#034; height=&#034;35&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P(E_2) = |c_2|^2 = 0}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P(E_2) = |c_2|^2 = 0}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Esto ocurre porque la funci&#243;n de onda es una combinaci&#243;n de estados estacionarios impares, como has calculado en el segundo apartado del problema.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Niveles del oscilador arm&#243;nico cu&#225;ntico y diferencia de energ&#237;a entre dos estados (8397)</title>
		<link>https://ejercicios-fyq.com/Niveles-del-oscilador-armonico-cuantico-y-diferencia-de-energia-entre-dos</link>
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		<dc:date>2025-02-20T03:34:52Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Oscilador arm&#243;nico cu&#225;ntico</dc:subject>

		<description>
&lt;p&gt;Un oscilador arm&#243;nico cu&#225;ntico tiene un hamiltoniano dado por: &lt;br class='autobr' /&gt; &lt;br class='autobr' /&gt;
donde &#171;p&#187; es el operador momento, &#171;m&#187; es la masa de la part&#237;cula, es la frecuencia angular del oscilador, y &#171;x&#187; es el operador posici&#243;n. &lt;br class='autobr' /&gt;
a) Demuestra que los niveles de energ&#237;a del oscilador arm&#243;nico cu&#225;ntico son: &lt;br class='autobr' /&gt; b) Calcula la diferencia de energ&#237;a entre el primer estado excitado (n = 1) y el estado fundamental (n = 0).&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Fisica-cuantica-288" rel="directory"&gt;F&#237;sica cu&#225;ntica&lt;/a&gt;

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&lt;a href="https://ejercicios-fyq.com/Oscilador-armonico-cuantico" rel="tag"&gt;Oscilador arm&#243;nico cu&#225;ntico&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Un oscilador arm&#243;nico cu&#225;ntico tiene un hamiltoniano dado por:&lt;/p&gt;
&lt;p&gt;
&lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L181xH48/d78bc1351273d7b262593f5597a27d1d-154ec.png?1740022639' style='vertical-align:middle;' width='181' height='48' alt=&#034;H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2&#034; title=&#034;H = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2&#034; /&gt;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;donde &#171;p&#187; es el operador momento, &#171;m&#187; es la masa de la part&#237;cula, &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L18xH30/260b57b4fdee8c5a001c09b555ccd28d-fbe90.png?1732988599' style='vertical-align:middle;' width='18' height='30' alt=&#034;\omega&#034; title=&#034;\omega&#034; /&gt; es la frecuencia angular del oscilador, y &#171;x&#187; es el operador posici&#243;n.&lt;/p&gt;
&lt;p&gt;a) Demuestra que los niveles de energ&#237;a del oscilador arm&#243;nico cu&#225;ntico son:&lt;/p&gt;
&lt;p&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L326xH52/e1ecfe7c765052cee1bb71b647c2db3f-1e796.png?1740022639' style='vertical-align:middle;' width='326' height='52' alt=&#034;E_n = \hbar \omega \left(n + \frac{1}{2}\right), \quad n = 0, 1, 2, \dots&#034; title=&#034;E_n = \hbar \omega \left(n + \frac{1}{2}\right), \quad n = 0, 1, 2, \dots&#034; /&gt;&lt;/p&gt;
&lt;p&gt;b) Calcula la diferencia de energ&#237;a entre el primer estado excitado (n = 1) y el estado fundamental (n = 0).&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) Puedes escribir el hamiltoniano del oscilador arm&#243;nico en funci&#243;n de los operadores de creaci&#243;n (&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/cc5f97acddfc0a1713b7d66a425e1baf.png' style=&#034;vertical-align:middle;&#034; width=&#034;17&#034; height=&#034;20&#034; alt=&#034;a^\dagger&#034; title=&#034;a^\dagger&#034; /&gt;) y aniquilaci&#243;n (&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0cc175b9c0f1b6a831c399e269772661.png' style=&#034;vertical-align:middle;&#034; width=&#034;10&#034; height=&#034;10&#034; alt=&#034;a&#034; title=&#034;a&#034; /&gt;), cuyas f&#243;rmulas son: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d1a072c4dcd966ab90a3faebcca0c325.png' style=&#034;vertical-align:middle;&#034; width=&#034;214&#034; height=&#034;54&#034; alt=&#034;a = \sqrt{\frac{m\omega}{2\hbar}} \left(x + \frac{i}{m\omega} p\right)&#034; title=&#034;a = \sqrt{\frac{m\omega}{2\hbar}} \left(x + \frac{i}{m\omega} p\right)&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/8a4f211936b60bce8a0babe4c13a533c.png' style=&#034;vertical-align:middle;&#034; width=&#034;223&#034; height=&#034;54&#034; alt=&#034;\quad a^\dagger = \sqrt{\frac{m\omega}{2\hbar}} \left(x - \frac{i}{m\omega} p\right)&#034; title=&#034;\quad a^\dagger = \sqrt{\frac{m\omega}{2\hbar}} \left(x - \frac{i}{m\omega} p\right)&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La f&#243;rmula del hamiltoniano es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/afcd0fd5d72ca8ce26c90083cda9fbb5.png' style=&#034;vertical-align:middle;&#034; width=&#034;194&#034; height=&#034;52&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{H = \hbar \omega \left(a^\dagger a + \frac{1}{2}\right)}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{H = \hbar \omega \left(a^\dagger a + \frac{1}{2}\right)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El operador n&#250;mero (&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/4f9469897f41b112a3c27d1b05088abe.png' style=&#034;vertical-align:middle;&#034; width=&#034;77&#034; height=&#034;20&#034; alt=&#034;N = a^\dagger a&#034; title=&#034;N = a^\dagger a&#034; /&gt;) cumple con la condici&#243;n (&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e4d0f745d1aeb9ab04965b4a444ead6c.png' style=&#034;vertical-align:middle;&#034; width=&#034;111&#034; height=&#034;23&#034; alt=&#034;N |n\rangle = n |n\rangle&#034; title=&#034;N |n\rangle = n |n\rangle&#034; /&gt;), donde &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/dc42978c67f1b8fc875820171cb2c937.png' style=&#034;vertical-align:middle;&#034; width=&#034;22&#034; height=&#034;23&#034; alt=&#034;|n\rangle&#034; title=&#034;|n\rangle&#034; /&gt; son los estados propios del oscilador arm&#243;nico. Puedes reescribir la ecuaci&#243;n anterior como: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a209ac3a2b2eb711f9fc12a17ba9132b.png' style=&#034;vertical-align:middle;&#034; width=&#034;181&#034; height=&#034;52&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{H = \hbar \omega \left(N + \frac{1}{2}\right)}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{H = \hbar \omega \left(N + \frac{1}{2}\right)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si das los distintos valores al operador n&#250;mero tienes la ecuaci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/bb61a0bd629aa8d9054bf7661a419dd1.png' style=&#034;vertical-align:middle;&#034; width=&#034;353&#034; height=&#034;40&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{E_n = \hbar \omega \left(n + \frac{1}{2}\right), \quad n = 0, 1, 2, {...}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{E_n = \hbar \omega \left(n + \frac{1}{2}\right), \quad n = 0, 1, 2, {...}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) Lo primero que debes hacer es escribir las ecuaciones para las energ&#237;as del estado fundamental y el primer estado excitado: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/eac66746a1aef6cbd17a169a638b6eb4.png' style=&#034;vertical-align:middle;&#034; width=&#034;227&#034; height=&#034;65&#034; alt=&#034;\left E_0 = \hbar \omega \left(0 + \frac{1}{2}\right) = {\color[RGB]{0,112,192}{\bm{\frac{\hbar \omega}{2}}}} \atop E_1 = \hbar \omega \left(1 + \frac{1}{2}\right) = {\color[RGB]{0,112,192}{\bm{\frac{3\hbar \omega}{2}}}} \right \}&#034; title=&#034;\left E_0 = \hbar \omega \left(0 + \frac{1}{2}\right) = {\color[RGB]{0,112,192}{\bm{\frac{\hbar \omega}{2}}}} \atop E_1 = \hbar \omega \left(1 + \frac{1}{2}\right) = {\color[RGB]{0,112,192}{\bm{\frac{3\hbar \omega}{2}}}} \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La diferencia de energ&#237;a es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/110dc42c9e7ec324a6d597cfe40261ef.png' style=&#034;vertical-align:middle;&#034; width=&#034;421&#034; height=&#034;45&#034; alt=&#034;\Delta E = E_1 - E_0 = \frac{3\hbar \omega}{2} - \frac{\hbar \omega}{2}\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{\Delta E = \hbar \omega}}}&#034; title=&#034;\Delta E = E_1 - E_0 = \frac{3\hbar \omega}{2} - \frac{\hbar \omega}{2}\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{\Delta E = \hbar \omega}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Constante de normalizaci&#243;n y probabilidad de estar en el estado fundamental sabiendo la funci&#243;n de onda (8396)</title>
		<link>https://ejercicios-fyq.com/Constante-de-normalizacion-y-probabilidad-de-estar-en-el-estado-fundamental</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Constante-de-normalizacion-y-probabilidad-de-estar-en-el-estado-fundamental</guid>
		<dc:date>2025-02-12T04:31:53Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Funci&#243;n de onda</dc:subject>
		<dc:subject>Constante normalizaci&#243;n</dc:subject>

		<description>
&lt;p&gt;Una part&#237;cula de masa &#171;m&#187; est&#225; confinada en una caja unidimensional de longitud &#171;L&#187;, con paredes infinitamente altas, es decir, con potencial infinito fuera de la caja. La funci&#243;n de onda inicial de la part&#237;cula es: &lt;br class='autobr' /&gt; &lt;br class='autobr' /&gt;
a) Determina la constante de normalizaci&#243;n &#171;A&#187;. &lt;br class='autobr' /&gt;
b) Encuentra la probabilidad de que la part&#237;cula se encuentre en el estado fundamental (n=1).&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Funcion-de-onda" rel="tag"&gt;Funci&#243;n de onda&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Constante-normalizacion" rel="tag"&gt;Constante normalizaci&#243;n&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Una part&#237;cula de masa &#171;m&#187; est&#225; confinada en una caja unidimensional de longitud &#171;L&#187;, con paredes infinitamente altas, es decir, con potencial infinito fuera de la caja. La funci&#243;n de onda inicial de la part&#237;cula es:&lt;/p&gt;
&lt;p&gt;
&lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L335xH53/5198c5a1d37d7b6573bf695c9eac185c-63e92.png?1739335375' style='vertical-align:middle;' width='335' height='53' alt=&#034;\Psi(x,0) = \left\{ {A\cdot sen(\frac{\pi\cdot x}{L}),\ \ 0\leq x \leq L \atop 0,\ \ \ \ \ \text{en~otro~caso}}&#034; title=&#034;\Psi(x,0) = \left\{ {A\cdot sen(\frac{\pi\cdot x}{L}),\ \ 0\leq x \leq L \atop 0,\ \ \ \ \ \text{en~otro~caso}}&#034; /&gt;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;a) Determina la constante de normalizaci&#243;n &#171;A&#187;.&lt;/p&gt;
&lt;p&gt;b) Encuentra la probabilidad de que la part&#237;cula se encuentre en el estado fundamental (n=1).&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) La normalizaci&#243;n de la funci&#243;n de onda debe cumplir: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/4b24279652ae337bcf461da64ec482c9.png' style=&#034;vertical-align:middle;&#034; width=&#034;181&#034; height=&#034;54&#034; alt=&#034;\int_0^L |\Psi(x,0)|^2 dx = 1&#034; title=&#034;\int_0^L |\Psi(x,0)|^2 dx = 1&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El cuadrado de la funci&#243;n de onda, en el intervalo de la integral, y la integral que tienes que resolver son: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6d4b186521e259f39ae53f5bb1133f82.png' style=&#034;vertical-align:middle;&#034; width=&#034;577&#034; height=&#034;54&#034; alt=&#034;{\color[RGB]{0,112,192}{\bm{|\Psi(x,0)|^2 = A^2 \sen^2 \left(\frac{\pi x}{L}\right)}}}\ \to\ {\color[RGB]{2,112,20}{\bm{\int_0^L A^2 \sen^2 \left(\frac{\pi x}{L}\right) dx = 1}}}&#034; title=&#034;{\color[RGB]{0,112,192}{\bm{|\Psi(x,0)|^2 = A^2 \sen^2 \left(\frac{\pi x}{L}\right)}}}\ \to\ {\color[RGB]{2,112,20}{\bm{\int_0^L A^2 \sen^2 \left(\frac{\pi x}{L}\right) dx = 1}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Se trata de una integral con integrando trignom&#233;trico y puedes resolverla si tienes en cuenta la ecuaci&#243;n del cuadrado del seno: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/951d7112f71ab61883ebe8719b7d454c.png' style=&#034;vertical-align:middle;&#034; width=&#034;220&#034; height=&#034;49&#034; alt=&#034;\color[RGB]{0,112,192}{\bm{\sen^2 \alpha = \frac{1 - \cos(2\alpha)}{2}}}&#034; title=&#034;\color[RGB]{0,112,192}{\bm{\sen^2 \alpha = \frac{1 - \cos(2\alpha)}{2}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Sustituyes en la funci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/b6f4c03fcec18892a8fa2a678067b3c6.png' style=&#034;vertical-align:middle;&#034; width=&#034;472&#034; height=&#034;54&#034; alt=&#034;\int_0^L A^2 \sen^2 \left(\frac{\pi x}{L}\right) dx = \frac{A^2}{2} \int_0^L \left[1 - \cos\left(\frac{2\pi x}{L}\right)}\right] dx&#034; title=&#034;\int_0^L A^2 \sen^2 \left(\frac{\pi x}{L}\right) dx = \frac{A^2}{2} \int_0^L \left[1 - \cos\left(\frac{2\pi x}{L}\right)}\right] dx&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Para hacer la integral la separas en dos integrales: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6e3e9843f3331f06d8158b125b8ce79b.png' style=&#034;vertical-align:middle;&#034; width=&#034;340&#034; height=&#034;54&#034; alt=&#034;A^2 \int_0^L \frac{1}{2} dx - A^2 \int_0^L \frac{\cos\left(\frac{2\pi x}{L}\right)}{2} dx = 1&#034; title=&#034;A^2 \int_0^L \frac{1}{2} dx - A^2 \int_0^L \frac{\cos\left(\frac{2\pi x}{L}\right)}{2} dx = 1&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La primera integral es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e39052681d884fd49e9cc8e5b4c7183b.png' style=&#034;vertical-align:middle;&#034; width=&#034;179&#034; height=&#034;54&#034; alt=&#034;A^2 \int_0^L \frac{1}{2} dx = \color[RGB]{0,112,192}{\bm{A^2 \frac{L}{2}}}&#034; title=&#034;A^2 \int_0^L \frac{1}{2} dx = \color[RGB]{0,112,192}{\bm{A^2 \frac{L}{2}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La integral trigonom&#233;trica es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/ae064947559cf17effada58003bbcbfc.png' style=&#034;vertical-align:middle;&#034; width=&#034;494&#034; height=&#034;58&#034; alt=&#034;\frac{A^2}{2} \int_0^L \cos\left(\frac{2\pi x}{L}\right) dx = \frac{A^2\cdot L}{4\pi}\cdot \left[\sen\left(\frac{2\pi\cdot x}{L}\right)\right]_0^L = \color[RGB]{0,112,192}{\bf 0}&#034; title=&#034;\frac{A^2}{2} \int_0^L \cos\left(\frac{2\pi x}{L}\right) dx = \frac{A^2\cdot L}{4\pi}\cdot \left[\sen\left(\frac{2\pi\cdot x}{L}\right)\right]_0^L = \color[RGB]{0,112,192}{\bf 0}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La constante de normalizaci&#243;n ser&#225;: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0cc35dd1edc5fadd9743fdf2435edea0.png' style=&#034;vertical-align:middle;&#034; width=&#034;243&#034; height=&#034;53&#034; alt=&#034;\frac{A^2\cdot L}{2} = 1\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{A = \sqrt{\frac{2}{L}}}}}&#034; title=&#034;\frac{A^2\cdot L}{2} = 1\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{A = \sqrt{\frac{2}{L}}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) La funci&#243;n de onda del estado fundamental de una part&#237;cula en una dimensi&#243;n es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/90328f377f7ba4d9b3a2dd855c1463b7.png' style=&#034;vertical-align:middle;&#034; width=&#034;206&#034; height=&#034;52&#034; alt=&#034;\psi_1 = \sqrt{\frac{2}{L}}\cdot \sen\left(\frac{\pi\cdot x}{L}\right)&#034; title=&#034;\psi_1 = \sqrt{\frac{2}{L}}\cdot \sen\left(\frac{\pi\cdot x}{L}\right)&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Observa que es de la misma forma que la funci&#243;n de onda y eso va a ser muy relevante. La probabilidad de que est&#233; en el estado fundamental ser&#225;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c91d1d64fe852f06e524328d4855d9fa.png' style=&#034;vertical-align:middle;&#034; width=&#034;742&#034; height=&#034;70&#034; alt=&#034;{\color[RGB]{2,112,20}{\bm{P_1 = \left| \int_0^L \psi_1^*(x) \Psi(x,0) dx \right|^2}}}\ \to\ P_1 = \left| \int_0^L \sqrt{\frac{2}{L}} \sen\left(\frac{\pi x}{L}\right) \cdot \sqrt{\frac{2}{L}} \sen\left(\frac{\pi x}{L}\right) dx \right|^2&#034; title=&#034;{\color[RGB]{2,112,20}{\bm{P_1 = \left| \int_0^L \psi_1^*(x) \Psi(x,0) dx \right|^2}}}\ \to\ P_1 = \left| \int_0^L \sqrt{\frac{2}{L}} \sen\left(\frac{\pi x}{L}\right) \cdot \sqrt{\frac{2}{L}} \sen\left(\frac{\pi x}{L}\right) dx \right|^2&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Simplificas la expresi&#243;n anterior y resuelves: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/33bdabdebd5c1ada9ace8c21efb00719.png' style=&#034;vertical-align:middle;&#034; width=&#034;463&#034; height=&#034;59&#034; alt=&#034;P_1 = \left| \int_0^L \frac{2}{L} \sin^2\left(\frac{\pi x}{L}\right) dx \right|^2 = \frac{2}{L}\cdot {\color[RGB]{0,112,192}{\bm{\frac{L}{2}}}}\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{P_1 = 1}}}&#034; title=&#034;P_1 = \left| \int_0^L \frac{2}{L} \sin^2\left(\frac{\pi x}{L}\right) dx \right|^2 = \frac{2}{L}\cdot {\color[RGB]{0,112,192}{\bm{\frac{L}{2}}}}\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{P_1 = 1}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Frecuencia umbral y potencial de extracci&#243;n de un metal con los datos de un experimento (8211)</title>
		<link>https://ejercicios-fyq.com/Frecuencia-umbral-y-potencial-de-extraccion-de-un-metal-con-los-datos-de-un</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Frecuencia-umbral-y-potencial-de-extraccion-de-un-metal-con-los-datos-de-un</guid>
		<dc:date>2024-05-16T04:54:09Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Efecto fotoel&#233;ctrico</dc:subject>
		<dc:subject>RESUELTO</dc:subject>

		<description>
&lt;p&gt;Cuando se ilumina una cierta superficie met&#225;lica con luz de diferentes longitudes de onda y se miden los potenciales que detienen los fotoelectrones, se obtienen los valores que se muestran en la siguiente tabla: &lt;br class='autobr' /&gt; &lt;br class='autobr' /&gt;
Representando el potencial en funci&#243;n de la frecuencia, determina: &lt;br class='autobr' /&gt;
a) La frecuencia umbral. &lt;br class='autobr' /&gt;
b) El trabajo de extracci&#243;n del metal. &lt;br class='autobr' /&gt;
c) La constante de Planck y el error relativo cometido.&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Efecto-fotoelectrico-305" rel="tag"&gt;Efecto fotoel&#233;ctrico&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Cuando se ilumina una cierta superficie met&#225;lica con luz de diferentes longitudes de onda y se miden los potenciales que detienen los fotoelectrones, se obtienen los valores que se muestran en la siguiente tabla:&lt;/p&gt;
&lt;p&gt;
&lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L480xH55/c8c9b4a8c60e9af8640a66ac4683ef80-b4581.png?1732979135' style='vertical-align:middle;' width='480' height='55' alt=&#034;\begin{tabular}{| c | c | c | c | c | c | c |} \hline \lambda\ (10^{-7}\ m)&amp;3.66&amp;4.05&amp;4.36&amp;4.92&amp;5.46&amp;5.79 \\\hline \Delta V\ (V)&amp;1.48&amp;1.15&amp;0.93&amp;0.62&amp;0.36&amp;0.24 \\\hline \end{tabular}&#034; title=&#034;\begin{tabular}{| c | c | c | c | c | c | c |} \hline \lambda\ (10^{-7}\ m)&amp;3.66&amp;4.05&amp;4.36&amp;4.92&amp;5.46&amp;5.79 \\\hline \Delta V\ (V)&amp;1.48&amp;1.15&amp;0.93&amp;0.62&amp;0.36&amp;0.24 \\\hline \end{tabular}&#034; /&gt;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;Representando el potencial en funci&#243;n de la frecuencia, determina:&lt;/p&gt;
&lt;p&gt;a) La frecuencia umbral.&lt;/p&gt;
&lt;p&gt;b) El trabajo de extracci&#243;n del metal.&lt;/p&gt;
&lt;p&gt;c) La constante de Planck y el error relativo cometido.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Lo primero que debes hacer, dado que te indica el problema que tienes que representar el potencial en funci&#243;n de la frecuencia, es convertir los datos de longitud de onda a frecuencia. Si tienes en cuenta la relaci&#243;n entre ambas magnitudes y la velocidad de propagaci&#243;n, puedes despejar el valor de la frecuencia: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/3b3cafe5e6a8e3ff3b9620c0d0982723.png' style=&#034;vertical-align:middle;&#034; width=&#034;187&#034; height=&#034;44&#034; alt=&#034;c = \lambda\cdot \nu\ \to\ \color[RGB]{2,112,20}{\bm{\nu = \frac{c}{\lambda}}}&#034; title=&#034;c = \lambda\cdot \nu\ \to\ \color[RGB]{2,112,20}{\bm{\nu = \frac{c}{\lambda}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si divides el valor de la velocidad de propagaci&#243;n de luz en el vac&#237;o entre cada valor de longitud de onda, obtienes una tabla como la que sigue, que ser&#225; la que representes: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/68b9c93baaa7f8538f4a495ac178ee0a.png' style=&#034;vertical-align:middle;&#034; width=&#034;487&#034; height=&#034;55&#034; alt=&#034;\begin{tabular}{| c | c | c | c | c | c | c |} \hline \nu\ (10^{14}\ Hz)&amp;8.19&amp;7.41&amp;6.88&amp;6.10&amp;5.49&amp;5.18 \\\hline \Delta V\ (V)&amp;1.48&amp;1.15&amp;0.93&amp;0.62&amp;0.36&amp;0.24 \\\hline \end{tabular}&#034; title=&#034;\begin{tabular}{| c | c | c | c | c | c | c |} \hline \nu\ (10^{14}\ Hz)&amp;8.19&amp;7.41&amp;6.88&amp;6.10&amp;5.49&amp;5.18 \\\hline \Delta V\ (V)&amp;1.48&amp;1.15&amp;0.93&amp;0.62&amp;0.36&amp;0.24 \\\hline \end{tabular}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; La representaci&#243;n gr&#225;fica y el ajuste lineal de la misma da lugar a: &lt;br/&gt;&lt;/p&gt;
&lt;div class='spip_document_1991 spip_document spip_documents spip_document_image spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt; &lt;a href='https://ejercicios-fyq.com/IMG/png/ej_8211.png' class=&#034;spip_doc_lien mediabox&#034; type=&#034;image/png&#034;&gt; &lt;img src='https://ejercicios-fyq.com/IMG/png/ej_8211.png' width=&#034;1812&#034; height=&#034;950&#034; alt='' /&gt;&lt;/a&gt;
&lt;/figure&gt;
&lt;/div&gt; &lt;p&gt;&lt;br/&gt; Si clicas en la imagen podr&#225;s ver el gr&#225;fico con m&#225;s detalle. &lt;br/&gt; &lt;br/&gt; La energ&#237;a de los fotoelectrones es la diferencia entre la energ&#237;a de la radiaci&#243;n y la energ&#237;a umbral, pero esta energ&#237;a es la misma que la energ&#237;a el&#233;ctrica que se usa para detenerlos. Puedes igualar ambas expresiones y obtener: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/b5d35c798541211ea9197ac2578f5b3a.png' style=&#034;vertical-align:middle;&#034; width=&#034;402&#034; height=&#034;52&#034; alt=&#034;\left E = h\cdot \nu - h\cdot \nu_0 \atop E = \Delta\cdot e \right \}\ \to\ \color[RGB]{2,112,20}{\bm{\Delta V = \frac{h}{e}\cdot \nu - \frac{E_0}{e}}}&#034; title=&#034;\left E = h\cdot \nu - h\cdot \nu_0 \atop E = \Delta\cdot e \right \}\ \to\ \color[RGB]{2,112,20}{\bm{\Delta V = \frac{h}{e}\cdot \nu - \frac{E_0}{e}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Por semejanza entre la ecuaci&#243;n de la recta de la gr&#225;fica y la ecuaci&#243;n anterior, puedes deducir que: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/35c2c0a8e0ccb9e63c3888c786dd026a.png' style=&#034;vertical-align:middle;&#034; width=&#034;194&#034; height=&#034;95&#034; alt=&#034;\left \dfrac{h}{e} = 4.1\cdot 10^{-15}\ \dfrac{V}{Hz} \atop \dfrac{E_0}{e} = 1.89\ V\right \}&#034; title=&#034;\left \dfrac{h}{e} = 4.1\cdot 10^{-15}\ \dfrac{V}{Hz} \atop \dfrac{E_0}{e} = 1.89\ V\right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; a) La frecuencia umbral la obtienes de la ordenada en el origen, teniendo en cuenta que la energ&#237;a umbral la puedes escribir en funci&#243;n de la frencuencia umbral: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c5173b5fe44389bc4f84aa2a1ccfbc08.png' style=&#034;vertical-align:middle;&#034; width=&#034;319&#034; height=&#034;55&#034; alt=&#034;\frac{E_0}{e} = \frac{h\cdot \nu_0}{e} = 1.89\ \to\ \color[RGB]{2,112,20}{\bm{\nu_0 = \frac{1.89}{\frac{h}{e}}}}&#034; title=&#034;\frac{E_0}{e} = \frac{h\cdot \nu_0}{e} = 1.89\ \to\ \color[RGB]{2,112,20}{\bm{\nu_0 = \frac{1.89}{\frac{h}{e}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Sustituyes los valores conocidos y obtienes la frecuencia umbral: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/3b5229ebe2c178b9ca2ef5aa327b1365.png' style=&#034;vertical-align:middle;&#034; width=&#034;361&#034; height=&#034;59&#034; alt=&#034;\nu_0 = \frac{1.89\ \cancel{V}}{4.1\cdot 10^{-15}\ \frac{\cancel{V}}{Hz}} = \fbox{\color[RGB]{192,0,0}{\bm{4.61\cdot 10^{14}\ Hz}}}&#034; title=&#034;\nu_0 = \frac{1.89\ \cancel{V}}{4.1\cdot 10^{-15}\ \frac{\cancel{V}}{Hz}} = \fbox{\color[RGB]{192,0,0}{\bm{4.61\cdot 10^{14}\ Hz}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) El trabajo de extracci&#243;n es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/9decd660640c4baaadbe27438d7612ee.png' style=&#034;vertical-align:middle;&#034; width=&#034;625&#034; height=&#034;46&#034; alt=&#034;V_0 = \frac{E_0}{e}\ \to\ E_0 = V_0\cdot e = 1.89\ V\cdot 1.6\cdot 10^{-19}\ C = \fbox{\color[RGB]{192,0,0}{\bm{3.02\cdot 10^{-19}\ J}}}&#034; title=&#034;V_0 = \frac{E_0}{e}\ \to\ E_0 = V_0\cdot e = 1.89\ V\cdot 1.6\cdot 10^{-19}\ C = \fbox{\color[RGB]{192,0,0}{\bm{3.02\cdot 10^{-19}\ J}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; c) La constante de Planck la obtienes de la pendiente de la recta representada: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/91c44bea338e7dbbc0fc82e6605447cb.png' style=&#034;vertical-align:middle;&#034; width=&#034;736&#034; height=&#034;45&#034; alt=&#034;\frac{h}{e} = 4.1\cdot 10^{-15}\ \frac{V}{Hz}\ \to\ h = 4.1\cdot 10^{-15}\ \frac{V}{Hz}\cdot 1.6\cdot 10^{-19}\ C = \fbox{\color[RGB]{192,0,0}{\bm{6.56\cdot 10^{-34}\ J\cdot s}}}&#034; title=&#034;\frac{h}{e} = 4.1\cdot 10^{-15}\ \frac{V}{Hz}\ \to\ h = 4.1\cdot 10^{-15}\ \frac{V}{Hz}\cdot 1.6\cdot 10^{-19}\ C = \fbox{\color[RGB]{192,0,0}{\bm{6.56\cdot 10^{-34}\ J\cdot s}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; El error de este valor experimental es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/dbdf5c94434b3b76839674ea55553248.png' style=&#034;vertical-align:middle;&#034; width=&#034;407&#034; height=&#034;50&#034; alt=&#034;\varepsilon_r = \frac{\bar h - h}{\bar h}\cdot 100 = \frac{6.63 - 6.56}{6.63}\cdot 100 = \fbox{\color[RGB]{192,0,0}{\bf 1\ \%}}&#034; title=&#034;\varepsilon_r = \frac{\bar h - h}{\bar h}\cdot 100 = \frac{6.63 - 6.56}{6.63}\cdot 100 = \fbox{\color[RGB]{192,0,0}{\bf 1\ \%}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Efecto fotoel&#233;ctrico: longitud de onda y frecuencia umbral para dos metales (8127)</title>
		<link>https://ejercicios-fyq.com/Efecto-fotoelectrico-longitud-de-onda-y-frecuencia-umbral-para-dos-metales-8127</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Efecto-fotoelectrico-longitud-de-onda-y-frecuencia-umbral-para-dos-metales-8127</guid>
		<dc:date>2024-01-25T04:37:28Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Efecto fotoel&#233;ctrico</dc:subject>
		<dc:subject>RESUELTO</dc:subject>

		<description>
&lt;p&gt;La funci&#243;n trabajo del K es 2.2 eV y la del Ni 5.0 eV. &lt;br class='autobr' /&gt;
a) Calcula las frecuencias y longitudes de onda umbral para estos dos metales. &lt;br class='autobr' /&gt;
b) &#191;Dar&#225; lugar la luz ultravioleta de longitud de onda 400 nm al efecto fotoel&#233;ctrico en el K? &#191;Y en el Ni? &lt;br class='autobr' /&gt;
c) Calcula la m&#225;xima energ&#237;a cin&#233;tica de los electrones emitidos en b).&lt;/p&gt;


-
&lt;a href="https://ejercicios-fyq.com/Fisica-cuantica-288" rel="directory"&gt;F&#237;sica cu&#225;ntica&lt;/a&gt;

/ 
&lt;a href="https://ejercicios-fyq.com/Efecto-fotoelectrico-305" rel="tag"&gt;Efecto fotoel&#233;ctrico&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;La funci&#243;n trabajo del K es 2.2 eV y la del Ni 5.0 eV.&lt;/p&gt;
&lt;p&gt;a) Calcula las frecuencias y longitudes de onda umbral para estos dos metales.&lt;/p&gt;
&lt;p&gt;b) &#191;Dar&#225; lugar la luz ultravioleta de longitud de onda 400 nm al efecto fotoel&#233;ctrico en el K? &#191;Y en el Ni?&lt;/p&gt;
&lt;p&gt;c) Calcula la m&#225;xima energ&#237;a cin&#233;tica de los electrones emitidos en b).&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Lo primero que puedes hacer es convertir los valores de funci&#243;n trabajo de ambos metales a unidad SI: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/2e3d6798240db463ff58df1af4b69105.png' style=&#034;vertical-align:middle;&#034; width=&#034;420&#034; height=&#034;48&#034; alt=&#034;W_{\ce{K}} = 2.2\ \cancel{eV}\cdot \frac{1.6\cdot 10^{-19}\ J}{1\ \cancel{eV}} = \color[RGB]{0,112,192}{\bm{3.52\cdot 10^{-19}\ J}}&#034; title=&#034;W_{\ce{K}} = 2.2\ \cancel{eV}\cdot \frac{1.6\cdot 10^{-19}\ J}{1\ \cancel{eV}} = \color[RGB]{0,112,192}{\bm{3.52\cdot 10^{-19}\ J}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/82aeffd2ea9a834b24d6509d050718c6.png' style=&#034;vertical-align:middle;&#034; width=&#034;393&#034; height=&#034;48&#034; alt=&#034;W_{\ce{Ni}} = 5.0\ \cancel{eV}\cdot \frac{1.6\cdot 10^{-19}\ J}{1\ \cancel{eV}} = \color[RGB]{0,112,192}{\bm{8\cdot 10^{-19}\ J}}&#034; title=&#034;W_{\ce{Ni}} = 5.0\ \cancel{eV}\cdot \frac{1.6\cdot 10^{-19}\ J}{1\ \cancel{eV}} = \color[RGB]{0,112,192}{\bm{8\cdot 10^{-19}\ J}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; a) La funci&#243;n trabajo est&#225; relacionada con la frecuencia umbral por medio de la ecuaci&#243;n de Planck: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/536f3197129dc348246fb04ccf6bfd4b.png' style=&#034;vertical-align:middle;&#034; width=&#034;110&#034; height=&#034;19&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{W = h\cdot \nu_u}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{W = h\cdot \nu_u}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Solo tienes que aplicarla para cada uno de los metales: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/5eafe2a73473a18d7e9aa4b04a2cd1c3.png' style=&#034;vertical-align:middle;&#034; width=&#034;456&#034; height=&#034;50&#034; alt=&#034;\nu_{\ce{K}} = \frac{W_{\ce{K}}}{h} = \frac{3.52\cdot 10^{-19}\ \cancel{J}}{6.63\cdot 10^{-34}\ \cancel{J}\cdot s} = \fbox{\color[RGB]{192,0,0}{\bm{5.31\cdot 10^{14}\ Hz}}}&#034; title=&#034;\nu_{\ce{K}} = \frac{W_{\ce{K}}}{h} = \frac{3.52\cdot 10^{-19}\ \cancel{J}}{6.63\cdot 10^{-34}\ \cancel{J}\cdot s} = \fbox{\color[RGB]{192,0,0}{\bm{5.31\cdot 10^{14}\ Hz}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/08b6c7ed6dbc6b9a88facd3f111d8cfc.png' style=&#034;vertical-align:middle;&#034; width=&#034;463&#034; height=&#034;50&#034; alt=&#034;\nu_{\ce{Ni}} = \frac{W_{\ce{Ni}}}{h} = \frac{8\cdot 10^{-19}\ \cancel{J}}{6.63\cdot 10^{-34}\ \cancel{J}\cdot s} = \fbox{\color[RGB]{192,0,0}{\bm{1.21\cdot 10^{15}\ Hz}}}&#034; title=&#034;\nu_{\ce{Ni}} = \frac{W_{\ce{Ni}}}{h} = \frac{8\cdot 10^{-19}\ \cancel{J}}{6.63\cdot 10^{-34}\ \cancel{J}\cdot s} = \fbox{\color[RGB]{192,0,0}{\bm{1.21\cdot 10^{15}\ Hz}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; La longitud se onda se relaciona con la frecuencia por medio de la velocidad de propagaci&#243;n de la radiaci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/4d018b91b9a0bd1639825d70777b326b.png' style=&#034;vertical-align:middle;&#034; width=&#034;187&#034; height=&#034;43&#034; alt=&#034;c = \lambda\cdot \nu\ \to\ \color[RGB]{2,112,20}{\bm{\lambda = \frac{c}{\nu}}}&#034; title=&#034;c = \lambda\cdot \nu\ \to\ \color[RGB]{2,112,20}{\bm{\lambda = \frac{c}{\nu}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Los valores de las longitudes de onda son: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/200c272537ffce32ac69c1c53b56253c.png' style=&#034;vertical-align:middle;&#034; width=&#034;420&#034; height=&#034;50&#034; alt=&#034;\lambda_{\ce{K}} = \frac{c}{\nu_{\ce{K}}} = \frac{3\cdot 10^8\ m\cdot \cancel{s^{-1}}}{5.31\cdot 10^{14}\ \cancel{s^{-1}}} = \fbox{\color[RGB]{192,0,0}{\bm{5.65\cdot 10^{-7}\ m}}}&#034; title=&#034;\lambda_{\ce{K}} = \frac{c}{\nu_{\ce{K}}} = \frac{3\cdot 10^8\ m\cdot \cancel{s^{-1}}}{5.31\cdot 10^{14}\ \cancel{s^{-1}}} = \fbox{\color[RGB]{192,0,0}{\bm{5.65\cdot 10^{-7}\ m}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/599c3d566ca2fc9c8703fe61cc71b6dd.png' style=&#034;vertical-align:middle;&#034; width=&#034;426&#034; height=&#034;50&#034; alt=&#034;\lambda_{\ce{Ni}} = \frac{c}{\nu_{\ce{Ni}}} = \frac{3\cdot 10^8\ m\cdot \cancel{s^{-1}}}{1.21\cdot 10^{15}\ \cancel{s^{-1}}} = \fbox{\color[RGB]{192,0,0}{\bm{2.48\cdot 10^{-7}\ m}}}&#034; title=&#034;\lambda_{\ce{Ni}} = \frac{c}{\nu_{\ce{Ni}}} = \frac{3\cdot 10^8\ m\cdot \cancel{s^{-1}}}{1.21\cdot 10^{15}\ \cancel{s^{-1}}} = \fbox{\color[RGB]{192,0,0}{\bm{2.48\cdot 10^{-7}\ m}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) Se dar&#225; efecto fotoel&#233;ctrico si la longitud de onda de la radiaci&#243;n empleada es menor que la de la funci&#243;n trabajo. Los 400 nm equivalen a &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/5ca18a06747cf24728df9dcf3a635e8c.png' style=&#034;vertical-align:middle;&#034; width=&#034;92&#034; height=&#034;47&#034; alt=&#034;4\cdot 10^{-7}\ m&#034; title=&#034;4\cdot 10^{-7}\ m&#034; /&gt;, por lo que, &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/4470a0aa193384b38ddecf6764c80829.png' style=&#034;vertical-align:middle;&#034; width=&#034;526&#034; height=&#034;32&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\textbf{se produce para el K y no se produce para el Ni}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\textbf{se produce para el K y no se produce para el Ni}}}&#034; /&gt;. &lt;br/&gt; &lt;br/&gt; c) Solo tienes que tener en cuenta los datos del potasio, que es quien produce efecto fotoel&#233;ctrico. Los electrones adquieren una energ&#237;a cin&#233;tica que es la diferencia entre la energ&#237;a de la radiaci&#243;n incidente y la funci&#243;n trabajo del K: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/f7af9c57b03eafc36ecfe84ef9f85706.png' style=&#034;vertical-align:middle;&#034; width=&#034;572&#034; height=&#034;52&#034; alt=&#034;E_C = E_i - W_{\ce{K}} = h\cdot \nu_i - h\cdot \nu_{\ce{K}}\ \to\ \color[RGB]{2,112,20}{\bm{E_C = h\cdot c\cdot \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_{\ce{K}}}\right)}}&#034; title=&#034;E_C = E_i - W_{\ce{K}} = h\cdot \nu_i - h\cdot \nu_{\ce{K}}\ \to\ \color[RGB]{2,112,20}{\bm{E_C = h\cdot c\cdot \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_{\ce{K}}}\right)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Solo tienes que sustituir y calcular: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/8a3694ec681a96ea97457be4936ed012.png' style=&#034;vertical-align:middle;&#034; width=&#034;769&#034; height=&#034;52&#034; alt=&#034;E_C = 6.63\cdot 10^{-34}\ J\cdot \cancel{s}\cdot 3\cdot 10^8\ \frac{\cancel{m}}{\cancel{s}}\left(\frac{1}{4\cdot 10^{-7}} - \frac{1}{5.65\cdot 10^{-7}}\right)\ \cancel{m^{-1}} = \fbox{\color[RGB]{192,0,0}{\bm{1.45\cdot 10^{-19}\ J}}}&#034; title=&#034;E_C = 6.63\cdot 10^{-34}\ J\cdot \cancel{s}\cdot 3\cdot 10^8\ \frac{\cancel{m}}{\cancel{s}}\left(\frac{1}{4\cdot 10^{-7}} - \frac{1}{5.65\cdot 10^{-7}}\right)\ \cancel{m^{-1}} = \fbox{\color[RGB]{192,0,0}{\bm{1.45\cdot 10^{-19}\ J}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Probabilidad de encontrar una part&#237;cula en una zona de una caja unidimensional (8103)</title>
		<link>https://ejercicios-fyq.com/Probabilidad-de-encontrar-una-particula-en-una-zona-de-una-caja-unidimensional</link>
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		<dc:date>2023-12-01T05:33:26Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Ecuaci&#243;n Schr&#246;dinger</dc:subject>

		<description>
&lt;p&gt;Una part&#237;cula se mueve en el interior de una caja unidimensional de longitud a y potencial infinito en sus extremos calcular la probabilidad de encontrar la part&#237;cula a del lado izquierdo de la caja y para que valor del estado cu&#225;ntico n es m&#225;xima esta probabilidad.&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Una part&#237;cula se mueve en el interior de una caja unidimensional de longitud &lt;i&gt;a&lt;/i&gt; y potencial infinito en sus extremos calcular la probabilidad de encontrar la part&#237;cula a &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L6xH21/8b72600e177e2cb1546d4cbc3074c364-7fc41.png?1733048689' style='vertical-align:middle;' width='6' height='21' alt=&#034;\textstyle{1\over 8}&#034; title=&#034;\textstyle{1\over 8}&#034; /&gt; del lado izquierdo de la caja y para que valor del estado cu&#225;ntico &lt;i&gt;n&lt;/i&gt; es m&#225;xima esta probabilidad.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Como nuestro problema es unidimensional, vamos a considerar solo la componente &#171;x&#187; del sistema y la posici&#243;n de la part&#237;cula tendr&#225; que variar entre los puntos 0 y &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/df9b21c88db63bfacd059d96a51d2904.png' style=&#034;vertical-align:middle;&#034; width=&#034;7&#034; height=&#034;18&#034; alt=&#034;\textstyle{a\over 8}&#034; title=&#034;\textstyle{a\over 8}&#034; /&gt;. La funci&#243;n de onda que vamos a considerar, por lo tanto, ser&#225;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/bf0bf03978f26a55ce5b9a7e7d0adfb5.png' style=&#034;vertical-align:middle;&#034; width=&#034;122&#034; height=&#034;43&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\sqrt{\frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\sqrt{\frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La probabilidad que debes calcular la integral del producto de la conjugada de la funci&#243;n de onda por la propia funci&#243;n de onda. En este caso, al ser unidireccional, coinciden ambas ecuaciones: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/8c59d2bc3ebe335d38f51586c2edacb0.png' style=&#034;vertical-align:middle;&#034; width=&#034;417&#034; height=&#034;43&#034; alt=&#034;P(0, \frac{a}{8}) = \int_0^{\frac{a}{8}}\psi^*\cdot \psi\cdot dx = \int_0^{\frac{a}{8}} \frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)\cdot \frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)&#034; title=&#034;P(0, \frac{a}{8}) = \int_0^{\frac{a}{8}}\psi^*\cdot \psi\cdot dx = \int_0^{\frac{a}{8}} \frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)\cdot \frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si multiplicas ambas funciones y sacas del integrando las constante, tienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/011e4194657d4d3b42d5a4dbb55ae97a.png' style=&#034;vertical-align:middle;&#034; width=&#034;267&#034; height=&#034;44&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{P(0, \frac{a}{8}) = \frac{2}{a}\int_0^{\frac{a}{8}} sen^2\left(\frac{n\pi x}{a}\right)\cdot dx}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{P(0, \frac{a}{8}) = \frac{2}{a}\int_0^{\frac{a}{8}} sen^2\left(\frac{n\pi x}{a}\right)\cdot dx}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si aplicas la siguiente igualdad trigonom&#233;trica puedes hacer la integral m&#225;s f&#225;cil: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/208974ff322c92446c792719f834abe0.png' style=&#034;vertical-align:middle;&#034; width=&#034;140&#034; height=&#034;34&#034; alt=&#034;sen^2 \alpha = \frac{1-cos\ 2\alpha}{2}&#034; title=&#034;sen^2 \alpha = \frac{1-cos\ 2\alpha}{2}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La integral anterior, al sacar la constante &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/26da16dcd83476d49b0255dcbae8635b.png' style=&#034;vertical-align:middle;&#034; width=&#034;12&#034; height=&#034;45&#034; alt=&#034;\textstyle{1\over 2}&#034; title=&#034;\textstyle{1\over 2}&#034; /&gt; y operar con la otra que estaba fuera del integrando, queda como: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/246647ba8bb3a6d8a0299325376edd35.png' style=&#034;vertical-align:middle;&#034; width=&#034;321&#034; height=&#034;44&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{P(0, \frac{a}{8}) = \frac{1}{a} \int_0^{\frac{a}{8}} \left[1 - cos\ \left(\frac{2n\pi x}{a}\right) \right]\cdot dx}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{P(0, \frac{a}{8}) = \frac{1}{a} \int_0^{\frac{a}{8}} \left[1 - cos\ \left(\frac{2n\pi x}{a}\right) \right]\cdot dx}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Puedes dividir la integral en dos integrales; una de ellas es inmediata y la otra debe ser resuelta por sustituci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/1187ad4d7640fdd21ef0df024ea35722.png' style=&#034;vertical-align:middle;&#034; width=&#034;335&#034; height=&#034;50&#034; alt=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{a}\left[ \int_0^{\frac{a}{8}} dx - \int_0^{\frac{a}{8}} cos\ \left(\dfrac{2n\pi x}{a}\right)\cdot dx\right]&#034; title=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{a}\left[ \int_0^{\frac{a}{8}} dx - \int_0^{\frac{a}{8}} cos\ \left(\dfrac{2n\pi x}{a}\right)\cdot dx\right]&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El resultado de las integrales que obtienes es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/af8a21137cf9622d64bf70d9fe52b9ff.png' style=&#034;vertical-align:middle;&#034; width=&#034;314&#034; height=&#034;40&#034; alt=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{a} \left[x |_0^{\frac{a}{8}} - \frac{a}{2n\pi}\cdot sen\ \left(\frac{2n\pi x}{a}\right) |_0^{\frac{a}{8}\right]&#034; title=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{a} \left[x |_0^{\frac{a}{8}} - \frac{a}{2n\pi}\cdot sen\ \left(\frac{2n\pi x}{a}\right) |_0^{\frac{a}{8}\right]&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Aplicas los l&#237;mites superior e inferior en cada caso y obtienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/642d3f91d6f6bebcac110f5ee95163da.png' style=&#034;vertical-align:middle;&#034; width=&#034;444&#034; height=&#034;38&#034; alt=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{\cancel{a}}\cdot \frac{\cancel{a}}{8} - \frac{1}{\cancel{a}}\cdot \frac{\cancel{a}}{2n\pi}\cdot sen\ \frac{2n\pi \cancel{a}}{8 \cancel{a}} = \color[RGB]{0,112,192}{\bm{\frac{1}{8} - \frac{1}{2n\pi}\cdot sen\ \frac{n\pi}{4}}}&#034; title=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{\cancel{a}}\cdot \frac{\cancel{a}}{8} - \frac{1}{\cancel{a}}\cdot \frac{\cancel{a}}{2n\pi}\cdot sen\ \frac{2n\pi \cancel{a}}{8 \cancel{a}} = \color[RGB]{0,112,192}{\bm{\frac{1}{8} - \frac{1}{2n\pi}\cdot sen\ \frac{n\pi}{4}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La soluci&#243;n depende de la funci&#243;n seno, con lo que el resultado no es &#250;nico y es necesario hacer el an&#225;lisis de los posibles valores de &#171;n&#187; para obtener las probabilidades. &lt;br/&gt; &lt;br/&gt; &lt;u&gt;Para valores impares de &#171;n&#187;&lt;/u&gt;. &lt;br/&gt; &lt;br/&gt; Haces dos divisiones de valores impares de &#171;n&#187;: &lt;br/&gt; &lt;br/&gt; n = 1, 3, 9, 11, 17, 19... &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/9d8a6a2dc1ee39fdb2237f7d85862654.png' style=&#034;vertical-align:middle;&#034; width=&#034;154&#034; height=&#034;34&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} - \frac{\sqrt{2}}{4n\pi}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} - \frac{\sqrt{2}}{4n\pi}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; n = 5, 7, 13, 15, 21, 23... &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a1e73dcd8a23f10b3b0c0e815ea4e59f.png' style=&#034;vertical-align:middle;&#034; width=&#034;154&#034; height=&#034;34&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} + \frac{\sqrt{2}}{4n\pi}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} + \frac{\sqrt{2}}{4n\pi}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;u&gt;Para valores pares de &#171;n&#187;&lt;/u&gt;. &lt;br/&gt; &lt;br/&gt; En este caso la divisi&#243;n la haces en tres grupos de valores: &lt;br/&gt; &lt;br/&gt; n = 2, 10, 18, 26... &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/548257a4a0fae783d70d4e7d28528b89.png' style=&#034;vertical-align:middle;&#034; width=&#034;154&#034; height=&#034;30&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} - \frac{1}{2n\pi}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} - \frac{1}{2n\pi}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; n = 4, 8, 12, 16, 20, 22... &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e6fa40cfbd9af3730b1886eb0ecd7935.png' style=&#034;vertical-align:middle;&#034; width=&#034;105&#034; height=&#034;30&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; n = 6, 14, 22, 30... &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/5eb96ffd8c03b7bbe6d495bdebb3e56b.png' style=&#034;vertical-align:middle;&#034; width=&#034;154&#034; height=&#034;30&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} + \frac{1}{2n\pi}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} + \frac{1}{2n\pi}}}}&#034; /&gt;&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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