{"id":245746,"date":"2025-07-06T12:07:01","date_gmt":"2025-07-06T12:07:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=245746"},"modified":"2025-07-06T12:07:04","modified_gmt":"2025-07-06T12:07:04","slug":"express-the-kinetic-energy-of-an-electron-in-terms-of-its-wavelength-using-the-de-broglie-equation","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/06\/express-the-kinetic-energy-of-an-electron-in-terms-of-its-wavelength-using-the-de-broglie-equation\/","title":{"rendered":"Express the kinetic energy of an electron in terms of its wavelength using the de Broglie equation."},"content":{"rendered":"\n<p>Express the kinetic energy of an electron in terms of its wavelength using the de Broglie equation.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To express the kinetic energy of an electron in terms of its wavelength, we can use the de Broglie equation and principles of classical mechanics.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: De Broglie Wavelength<\/h3>\n\n\n\n<p>The de Broglie equation relates the wavelength (\u03bb\\lambda\u03bb) of a particle to its momentum (ppp):\u03bb=hp\\lambda = \\frac{h}{p}\u03bb=ph\u200b<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>hhh is Planck&#8217;s constant (h=6.626\u00d710\u221234\u2009J\u22c5sh = 6.626 \\times 10^{-34} \\, \\text{J} \\cdot \\text{s}h=6.626\u00d710\u221234J\u22c5s)<\/li>\n\n\n\n<li>ppp is the momentum of the electron.<\/li>\n<\/ul>\n\n\n\n<p>Momentum is defined as the product of the mass (mmm) of the electron and its velocity (vvv):p=mvp = mvp=mv<\/p>\n\n\n\n<p>Thus, the de Broglie wavelength becomes:\u03bb=hmv\\lambda = \\frac{h}{mv}\u03bb=mvh\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Solving for Momentum<\/h3>\n\n\n\n<p>Rearranging this equation to solve for momentum:p=h\u03bbp = \\frac{h}{\\lambda}p=\u03bbh\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Kinetic Energy of an Electron<\/h3>\n\n\n\n<p>The classical kinetic energy (K.E.K.E.K.E.) of an electron is given by:K.E.=12mv2K.E. = \\frac{1}{2}mv^2K.E.=21\u200bmv2<\/p>\n\n\n\n<p>Using the relationship for momentum (p=mvp = mvp=mv), we can solve for vvv:v=pmv = \\frac{p}{m}v=mp\u200b<\/p>\n\n\n\n<p>Substituting this expression for velocity into the kinetic energy formula:K.E.=12m(pm)2=p22mK.E. = \\frac{1}{2}m\\left(\\frac{p}{m}\\right)^2 = \\frac{p^2}{2m}K.E.=21\u200bm(mp\u200b)2=2mp2\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Expressing Kinetic Energy in Terms of Wavelength<\/h3>\n\n\n\n<p>Now substitute the expression for momentum from the de Broglie equation (p=h\u03bbp = \\frac{h}{\\lambda}p=\u03bbh\u200b) into the equation for kinetic energy:K.E.=(h\u03bb)22mK.E. = \\frac{\\left(\\frac{h}{\\lambda}\\right)^2}{2m}K.E.=2m(\u03bbh\u200b)2\u200b<\/p>\n\n\n\n<p>Simplifying:K.E.=h22m\u03bb2K.E. = \\frac{h^2}{2m\\lambda^2}K.E.=2m\u03bb2h2\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<p>Thus, the kinetic energy of an electron in terms of its de Broglie wavelength is:K.E.=h22m\u03bb2K.E. = \\frac{h^2}{2m\\lambda^2}K.E.=2m\u03bb2h2\u200b<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>hhh is Planck&#8217;s constant,<\/li>\n\n\n\n<li>mmm is the mass of the electron,<\/li>\n\n\n\n<li>\u03bb\\lambda\u03bb is the de Broglie wavelength of the electron.<\/li>\n<\/ul>\n\n\n\n<p>This equation shows that the kinetic energy of an electron is inversely proportional to the square of its wavelength. The shorter the wavelength, the higher the kinetic energy of the electron.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-706.jpeg\" alt=\"\" class=\"wp-image-245748\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Express the kinetic energy of an electron in terms of its wavelength using the de Broglie equation. The Correct Answer and Explanation is: To express the kinetic energy of an electron in terms of its wavelength, we can use the de Broglie equation and principles of classical mechanics. Step 1: De Broglie Wavelength The de [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-245746","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245746","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=245746"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245746\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245746"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245746"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245746"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}