{"id":231901,"date":"2025-06-11T14:34:10","date_gmt":"2025-06-11T14:34:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231901"},"modified":"2025-06-11T14:34:12","modified_gmt":"2025-06-11T14:34:12","slug":"visible-green-emission-spectral-line-for-krypton-kr-occurs-at-wavelength-of-533-822-nm","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/visible-green-emission-spectral-line-for-krypton-kr-occurs-at-wavelength-of-533-822-nm\/","title":{"rendered":"Visible (green) emission spectral line for krypton (Kr) occurs at wavelength of 533.822 nm"},"content":{"rendered":"\n

Visible (green) emission spectral line for krypton (Kr) occurs at wavelength of 533.822 nm. What is the frequency (v) of this light? (Mind the units carefully. Give your answers to six significant figures.) What is the magnitude of the energy change associated with the emission of one mole of photons of light with this wavelength?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Step 1: Calculate the frequency (\u03bd)<\/strong><\/h3>\n\n\n\n

The frequency \u03bd\\nu\u03bd of light is related to its wavelength \u03bb\\lambda\u03bb by the speed of light ccc:\u03bd=c\u03bb\\nu = \\frac{c}{\\lambda}\u03bd=\u03bbc\u200b<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • c=3.00\u00d7108\u00a0m\/sc = 3.00 \\times 10^8 \\ \\text{m\/s}c=3.00\u00d7108\u00a0m\/s (speed of light),<\/li>\n\n\n\n
  • \u03bb=533.822\u00a0nm=533.822\u00d710\u22129\u00a0m\\lambda = 533.822 \\ \\text{nm} = 533.822 \\times 10^{-9} \\ \\text{m}\u03bb=533.822\u00a0nm=533.822\u00d710\u22129\u00a0m<\/li>\n<\/ul>\n\n\n\n

    \u03bd=3.00\u00d7108 m\/s533.822\u00d710\u22129 m=5.61883\u00d71014 Hz\\nu = \\frac{3.00 \\times 10^8 \\ \\text{m\/s}}{533.822 \\times 10^{-9} \\ \\text{m}} = 5.61883 \\times 10^{14} \\ \\text{Hz}\u03bd=533.822\u00d710\u22129 m3.00\u00d7108 m\/s\u200b=5.61883\u00d71014 Hz<\/p>\n\n\n\n

    Answer (frequency):<\/strong> 5.61883\u00d71014 Hz\\boxed{5.61883 \\times 10^{14} \\ \\text{Hz}}5.61883\u00d71014 Hz\u200b<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Calculate the energy of one photon<\/strong><\/h3>\n\n\n\n

    The energy EEE of a single photon is given by Planck\u2019s equation:E=h\u03bdE = h\\nuE=h\u03bd<\/p>\n\n\n\n

    Where:<\/p>\n\n\n\n

      \n
    • h=6.626\u00d710\u221234\u00a0J\\cdotpsh = 6.626 \\times 10^{-34} \\ \\text{J\u00b7s}h=6.626\u00d710\u221234\u00a0J\\cdotps (Planck\u2019s constant),<\/li>\n\n\n\n
    • \u03bd=5.61883\u00d71014\u00a0Hz\\nu = 5.61883 \\times 10^{14} \\ \\text{Hz}\u03bd=5.61883\u00d71014\u00a0Hz<\/li>\n<\/ul>\n\n\n\n

      E=(6.626\u00d710\u221234 J\\cdotps)\u00d7(5.61883\u00d71014 Hz)=3.72381\u00d710\u221219 JE = (6.626 \\times 10^{-34} \\ \\text{J\u00b7s}) \\times (5.61883 \\times 10^{14} \\ \\text{Hz}) = 3.72381 \\times 10^{-19} \\ \\text{J}E=(6.626\u00d710\u221234 J\\cdotps)\u00d7(5.61883\u00d71014 Hz)=3.72381\u00d710\u221219 J<\/p>\n\n\n\n

      \n\n\n\n

      Step 3: Calculate energy for one mole of photons<\/strong><\/h3>\n\n\n\n

      1 mole of photons contains Avogadro\u2019s number of photons:Emol=E\u00d7NA=(3.72381\u00d710\u221219 J)\u00d7(6.022\u00d71023)E_{\\text{mol}} = E \\times N_A = (3.72381 \\times 10^{-19} \\ \\text{J}) \\times (6.022 \\times 10^{23})Emol\u200b=E\u00d7NA\u200b=(3.72381\u00d710\u221219 J)\u00d7(6.022\u00d71023)Emol=224,250 J\/mol=2.24250\u00d7105 J\/molE_{\\text{mol}} = 224,250 \\ \\text{J\/mol} = \\boxed{2.24250 \\times 10^5 \\ \\text{J\/mol}}Emol\u200b=224,250 J\/mol=2.24250\u00d7105 J\/mol\u200b<\/p>\n\n\n\n

      \n\n\n\n

      Final Answers:<\/strong><\/h3>\n\n\n\n
        \n
      • Frequency (\u03bd):<\/strong> 5.61883\u00d71014\u00a0Hz\\boxed{5.61883 \\times 10^{14} \\ \\text{Hz}}5.61883\u00d71014\u00a0Hz\u200b<\/li>\n\n\n\n
      • Energy per mole:<\/strong> 2.24250\u00d7105\u00a0J\/mol\\boxed{2.24250 \\times 10^5 \\ \\text{J\/mol}}2.24250\u00d7105\u00a0J\/mol\u200b<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        Explanation <\/strong><\/h3>\n\n\n\n

        The electromagnetic spectrum consists of a wide range of wavelengths and frequencies, with visible light occupying only a narrow portion. Each color of visible light corresponds to a specific wavelength and frequency. In this problem, we analyze the green emission line of krypton gas, which occurs at a wavelength of 533.822 nanometers.<\/p>\n\n\n\n

        To determine the frequency of this light, we use the relationship between speed, frequency, and wavelength of light: c=\u03bb\u03bdc = \\lambda \\nuc=\u03bb\u03bd. Rearranging, we find frequency \u03bd=c\/\u03bb\\nu = c \/ \\lambda\u03bd=c\/\u03bb. Substituting the known values\u2014speed of light c=3.00\u00d7108 m\/sc = 3.00 \\times 10^8 \\ \\text{m\/s}c=3.00\u00d7108 m\/s and wavelength converted to meters\u2014we calculate the frequency to be approximately 5.61883\u00d71014 Hz5.61883 \\times 10^{14} \\ \\text{Hz}5.61883\u00d71014 Hz, which is in the range expected for visible green light.<\/p>\n\n\n\n

        Next, we calculate the energy of one photon of this light using Planck\u2019s equation: E=h\u03bdE = h\\nuE=h\u03bd, where hhh is Planck\u2019s constant. This yields an energy of about 3.72381\u00d710\u221219 J3.72381 \\times 10^{-19} \\ \\text{J}3.72381\u00d710\u221219 J per photon. While this is a small amount of energy on a per-photon basis, the total energy becomes significant when considering a mole of photons.<\/p>\n\n\n\n

        To find the total energy associated with one mole of these photons, we multiply the energy of a single photon by Avogadro\u2019s number, 6.022\u00d71023 mol\u221216.022 \\times 10^{23} \\ \\text{mol}^{-1}6.022\u00d71023 mol\u22121. This gives a total energy release of 2.24250\u00d7105 J\/mol2.24250 \\times 10^5 \\ \\text{J\/mol}2.24250\u00d7105 J\/mol, which represents the magnitude of the energy change due to emission of light at this wavelength.<\/p>\n\n\n\n

        This kind of calculation is essential in fields like atomic spectroscopy, quantum chemistry, and photophysics, where understanding the energy of photons helps explain atomic transitions and chemical behavior<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Visible (green) emission spectral line for krypton (Kr) occurs at wavelength of 533.822 nm. What is the frequency (v) of this light? (Mind the units carefully. Give your answers to six significant figures.) What is the magnitude of the energy change associated with the emission of one mole of photons of light with this wavelength? […]<\/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-231901","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231901","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=231901"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231901\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231901"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231901"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231901"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}