{"id":261761,"date":"2025-07-20T08:25:30","date_gmt":"2025-07-20T08:25:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=261761"},"modified":"2025-07-20T08:25:33","modified_gmt":"2025-07-20T08:25:33","slug":"sodium-22-na-22-is-radioactive-and-decays-by-beta-plus-decay-mode-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/20\/sodium-22-na-22-is-radioactive-and-decays-by-beta-plus-decay-mode-2\/","title":{"rendered":"Sodium-22 (Na-22) is radioactive and decays by beta plus decay mode."},"content":{"rendered":"\n

Sodium-22 (Na-22) is radioactive and decays by beta plus decay mode. The half-life is 2.603 years. (a) Calculate the binding energy per nucleon for sodium-22. (b) Write the reaction for the decay. (c) Calculate Q for the decay reaction. (d) What is the mass of a sample of sodium-22 that has a decay rate of 127,500 decays per second? (e) How long will it take until 99.9% of the sodium has decayed? (Z= 10) Neon-20 19.992 440 u Ne Neon-21 20.993 847 u Neon-22 21.991 386 u (Z = 11) Sodium-22 21.994 437 u Na Sodium-23 22.989 770 u Sodium-24 23.990 963 u (Z= 12) Magnesium-22 21.993 267 u Mg Magnesium-23 22.994 128 u Magnesium-24 23.985 042 u proton 1.007 276 470 u neutron 1.008 664 904 u electron 0.000 548 580 u<\/p>\n\n\n\n

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

Let’s break down each part of the problem step-by-step:<\/p>\n\n\n\n

(a) Binding Energy per Nucleon for Sodium-22<\/h3>\n\n\n\n

The binding energy per nucleon is the energy required to remove a nucleon from the nucleus. We calculate it by first finding the mass defect<\/strong> and then converting it to energy.<\/p>\n\n\n\n

Step 1: Find the mass defect for Na-22.<\/h4>\n\n\n\n

The mass defect is the difference between the mass of the nucleus and the sum of the masses of the individual nucleons (protons and neutrons).<\/p>\n\n\n\n

    \n
  • Mass of Na-22 nucleus = 21.994 437 u<\/li>\n\n\n\n
  • Mass of 11 protons = 11\u00d71.007276470\u2009u=11.07904117\u2009u11 \\times 1.007 276 470 \\, \\text{u} = 11.079 041 17 \\, \\text{u}11\u00d71.007276470u=11.07904117u<\/li>\n\n\n\n
  • Mass of 11 neutrons = 11\u00d71.008664904\u2009u=11.095314944\u2009u11 \\times 1.008 664 904 \\, \\text{u} = 11.095 314 944 \\, \\text{u}11\u00d71.008664904u=11.095314944u<\/li>\n<\/ul>\n\n\n\n

    Total mass of nucleons = 11.079 041 17 + 11.095 314 944 = 22.174 356 114 u<\/p>\n\n\n\n

    Mass defect = Total mass of nucleons – Mass of the nucleus
    Mass defect = 22.174 356 114 u – 21.994 437 437 u = 0.179 918 677 u<\/p>\n\n\n\n

    Step 2: Convert the mass defect into energy.<\/h4>\n\n\n\n

    Energy E=\u0394m\u22c5c2E = \\Delta m \\cdot c^2E=\u0394m\u22c5c2 (Einstein\u2019s equation), but since we\u2019re working with atomic mass units, we use the conversion factor:<\/p>\n\n\n\n

    1 atomic mass unit (u) = 931.494 MeV<\/p>\n\n\n\n

    Energy = 0.179918677\u2009u\u00d7931.494\u2009MeV\/u=167.78\u2009MeV0.179 918 677 \\, \\text{u} \\times 931.494 \\, \\text{MeV\/u} = 167.78 \\, \\text{MeV}0.179918677u\u00d7931.494MeV\/u=167.78MeV<\/p>\n\n\n\n

    Step 3: Calculate binding energy per nucleon.<\/h4>\n\n\n\n

    Na-22 has 22 nucleons (11 protons + 11 neutrons), so:<\/p>\n\n\n\n

    Binding energy per nucleon = 167.78\u2009MeV22=7.63\u2009MeV\\frac{167.78 \\, \\text{MeV}}{22} = 7.63 \\, \\text{MeV}22167.78MeV\u200b=7.63MeV<\/p>\n\n\n\n

    Answer to part (a):<\/strong> The binding energy per nucleon for sodium-22 is 7.63 MeV<\/strong>.<\/p>\n\n\n\n

    \n\n\n\n

    (b) Write the Reaction for Beta Plus Decay of Sodium-22<\/h3>\n\n\n\n

    Beta plus decay (\u03b2+ decay) occurs when a proton is converted into a neutron, releasing a positron and a neutrino. For Na-22:Na1122\u2192Ne1022+\u03b2++\u03bde\\text{Na}^{22}_{11} \\rightarrow \\text{Ne}^{22}_{10} + \\beta^+ + \\nu_eNa1122\u200b\u2192Ne1022\u200b+\u03b2++\u03bde\u200b<\/p>\n\n\n\n

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

      \n
    • Na-22 is sodium-22.<\/li>\n\n\n\n
    • Ne-22 is neon-22.<\/li>\n\n\n\n
    • \u03b2+\\beta^+\u03b2+ is a positron.<\/li>\n\n\n\n
    • \u03bde\\nu_e\u03bde\u200b is the neutrino.<\/li>\n<\/ul>\n\n\n\n

      Answer to part (b):<\/strong> The decay reaction is:Na1122\u2192Ne1022+\u03b2++\u03bde\\text{Na}^{22}_{11} \\rightarrow \\text{Ne}^{22}_{10} + \\beta^+ + \\nu_eNa1122\u200b\u2192Ne1022\u200b+\u03b2++\u03bde\u200b<\/p>\n\n\n\n

      \n\n\n\n

      (c) Calculate Q for the Decay Reaction<\/h3>\n\n\n\n

      The Q-value of a decay is the difference in mass between the reactants and products, converted into energy. It can be calculated using:Q=(mNa-22\u2212mNe-22\u2212m\u03b2+\u2212m\u03bde)\u00d7c2Q = (m_{\\text{Na-22}} – m_{\\text{Ne-22}} – m_{\\beta^+} – m_{\\nu_e}) \\times c^2Q=(mNa-22\u200b\u2212mNe-22\u200b\u2212m\u03b2+\u200b\u2212m\u03bde\u200b\u200b)\u00d7c2<\/p>\n\n\n\n

      Step 1: Find the masses.<\/h4>\n\n\n\n
        \n
      • Mass of Na-22 = 21.994 437 u<\/li>\n\n\n\n
      • Mass of Ne-22 = 21.991 386 u<\/li>\n\n\n\n
      • Mass of positron (\u03b2+) = 0.000 548 580 u<\/li>\n\n\n\n
      • Mass of neutrino (\u03bd) is negligible, so we can ignore it.<\/li>\n<\/ul>\n\n\n\n

        Step 2: Calculate the Q-value.<\/h4>\n\n\n\n

        Q=(21.994437\u2009u\u221221.991386\u2009u\u22120.000548580\u2009u)\u00d7931.494\u2009MeV\/uQ = (21.994 437 \\, \\text{u} – 21.991 386 \\, \\text{u} – 0.000 548 580 \\, \\text{u}) \\times 931.494 \\, \\text{MeV\/u}Q=(21.994437u\u221221.991386u\u22120.000548580u)\u00d7931.494MeV\/uQ=(0.002502420\u2009u)\u00d7931.494\u2009MeV\/u=2.33\u2009MeVQ = (0.002 502 420 \\, \\text{u}) \\times 931.494 \\, \\text{MeV\/u} = 2.33 \\, \\text{MeV}Q=(0.002502420u)\u00d7931.494MeV\/u=2.33MeV<\/p>\n\n\n\n

        Answer to part (c):<\/strong> The Q-value for the decay is 2.33 MeV<\/strong>.<\/p>\n\n\n\n

        \n\n\n\n

        (d) Calculate the Mass of Sodium-22 for a Decay Rate of 127,500 Decays per Second<\/h3>\n\n\n\n

        We use the following formula to calculate the mass of Na-22 based on its decay rate:Decay rate=\u03bbN\\text{Decay rate} = \\lambda NDecay rate=\u03bbN<\/p>\n\n\n\n

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

          \n
        • \u03bb\\lambda\u03bb is the decay constant, \u03bb=ln\u20612t12\\lambda = \\frac{\\ln 2}{t_{\\frac{1}{2}}}\u03bb=t21\u200b\u200bln2\u200b<\/li>\n\n\n\n
        • NNN is the number of Na-22 nuclei.<\/li>\n<\/ul>\n\n\n\n

          Step 1: Calculate the decay constant.<\/h4>\n\n\n\n

          The half-life of Na-22 is 2.603 years. Convert this to seconds:t12=2.603\u2009years\u00d7365.25\u2009days\/year\u00d724\u2009hours\/day\u00d73600\u2009seconds\/hour=8.21\u00d7107\u2009secondst_{\\frac{1}{2}} = 2.603 \\, \\text{years} \\times 365.25 \\, \\text{days\/year} \\times 24 \\, \\text{hours\/day} \\times 3600 \\, \\text{seconds\/hour} = 8.21 \\times 10^7 \\, \\text{seconds}t21\u200b\u200b=2.603years\u00d7365.25days\/year\u00d724hours\/day\u00d73600seconds\/hour=8.21\u00d7107seconds<\/p>\n\n\n\n

          Now, calculate \u03bb\\lambda\u03bb:\u03bb=ln\u206128.21\u00d7107\u2009s=8.44\u00d710\u22129\u2009s\u22121\\lambda = \\frac{\\ln 2}{8.21 \\times 10^7 \\, \\text{s}} = 8.44 \\times 10^{-9} \\, \\text{s}^{-1}\u03bb=8.21\u00d7107sln2\u200b=8.44\u00d710\u22129s\u22121<\/p>\n\n\n\n

          Step 2: Find the number of Na-22 nuclei.<\/h4>\n\n\n\n

          N=Decay rate\u03bb=127,500\u2009decays\/s8.44\u00d710\u22129\u2009s\u22121=1.51\u00d71013\u2009nucleiN = \\frac{\\text{Decay rate}}{\\lambda} = \\frac{127,500 \\, \\text{decays\/s}}{8.44 \\times 10^{-9} \\, \\text{s}^{-1}} = 1.51 \\times 10^{13} \\, \\text{nuclei}N=\u03bbDecay rate\u200b=8.44\u00d710\u22129s\u22121127,500decays\/s\u200b=1.51\u00d71013nuclei<\/p>\n\n\n\n

          Step 3: Find the mass.<\/h4>\n\n\n\n

          The number of moles of Na-22 is:moles=NNA=1.51\u00d710136.022\u00d71023=2.51\u00d710\u221211\u2009moles\\text{moles} = \\frac{N}{N_A} = \\frac{1.51 \\times 10^{13}}{6.022 \\times 10^{23}} = 2.51 \\times 10^{-11} \\, \\text{moles}moles=NA\u200bN\u200b=6.022\u00d710231.51\u00d71013\u200b=2.51\u00d710\u221211moles<\/p>\n\n\n\n

          Now, convert this to mass:mass=moles\u00d7molar mass=2.51\u00d710\u221211\u2009mol\u00d721.994437\u2009g\/mol=5.52\u00d710\u221210\u2009g\\text{mass} = \\text{moles} \\times \\text{molar mass} = 2.51 \\times 10^{-11} \\, \\text{mol} \\times 21.994 437 \\, \\text{g\/mol} = 5.52 \\times 10^{-10} \\, \\text{g}mass=moles\u00d7molar mass=2.51\u00d710\u221211mol\u00d721.994437g\/mol=5.52\u00d710\u221210g<\/p>\n\n\n\n

          Answer to part (d):<\/strong> The mass of Na-22 is 5.52 \u00d7 10\u207b\u00b9\u2070 g<\/strong>.<\/p>\n\n\n\n

          \n\n\n\n

          (e) Time for 99.9% of Sodium-22 to Decay<\/h3>\n\n\n\n

          To find the time for 99.9% decay, we use the following formula based on the half-life:N(t)=N0e\u2212\u03bbtN(t) = N_0 e^{-\\lambda t}N(t)=N0\u200be\u2212\u03bbt<\/p>\n\n\n\n

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

            \n
          • N(t)N(t)N(t) is the amount remaining after time ttt.<\/li>\n\n\n\n
          • N0N_0N0\u200b is the initial amount.<\/li>\n\n\n\n
          • \u03bb\\lambda\u03bb is the decay constant.<\/li>\n\n\n\n
          • ttt is the time.<\/li>\n<\/ul>\n\n\n\n

            For 99.9% decay, 0.1% remains, so:0.001N0=N0e\u2212\u03bbt0.001 N_0 = N_0 e^{-\\lambda t}0.001N0\u200b=N0\u200be\u2212\u03bbt<\/p>\n\n\n\n

            Solve for ttt:e\u2212\u03bbt=0.001e^{-\\lambda t} = 0.001e\u2212\u03bbt=0.001\u2212\u03bbt=ln\u2061(0.001)-\\lambda t = \\ln(0.001)\u2212\u03bbt=ln(0.001)t=ln\u2061(0.001)\u2212\u03bb=\u22126.9078.44\u00d710\u22129\u2009s\u22121=8.19\u00d7105\u2009st = \\frac{\\ln(0.001)}{-\\lambda} = \\frac{-6.907}{8.44 \\times 10^{-9} \\, \\text{s}^{-1}} = 8.19 \\times 10^5 \\, \\text{s}t=\u2212\u03bbln(0.001)\u200b=8.44\u00d710\u22129s\u22121\u22126.907\u200b=8.19\u00d7105s<\/p>\n\n\n\n

            Convert to years:t=8.19\u00d7105\u2009s60\u00d760\u00d724\u00d7365.25\u224825.9\u2009yearst = \\frac{8.19 \\times 10^5 \\, \\text{s}}{60 \\times 60 \\times 24 \\times 365.25} \\approx 25.9 \\, \\text{years}t=60\u00d760\u00d724\u00d7365.258.19\u00d7105s\u200b\u224825.9years<\/p>\n\n\n\n

            Answer to part (e):<\/strong> It will take approximately 25.9 years<\/strong> for 99.9% of the sodium-22 to decay.<\/p>\n\n\n\n

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

            Sodium-22 (Na-22) is radioactive and decays by beta plus decay mode. The half-life is 2.603 years. (a) Calculate the binding energy per nucleon for sodium-22. (b) Write the reaction for the decay. (c) Calculate Q for the decay reaction. (d) What is the mass of a sample of sodium-22 that has a decay rate of […]<\/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-261761","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261761","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=261761"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261761\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=261761"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=261761"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=261761"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}