{"id":259968,"date":"2025-07-19T06:40:12","date_gmt":"2025-07-19T06:40:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=259968"},"modified":"2025-07-19T06:40:15","modified_gmt":"2025-07-19T06:40:15","slug":"sodium-22-na-22-is-radioactive-and-decays-by-beta-plus-decay-mode","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/19\/sodium-22-na-22-is-radioactive-and-decays-by-beta-plus-decay-mode\/","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 this down 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 amount of energy required to remove a nucleon (proton or neutron) from the nucleus, and it can be calculated using the mass defect and Einstein’s equation E=mc2E = mc^2E=mc2.<\/p>\n\n\n\n

    \n
  1. Mass of Na-22 nucleus<\/strong>: 21.994437 u<\/li>\n\n\n\n
  2. Mass of nucleons in Na-22<\/strong>: Sodium-22 consists of 11 protons and 11 neutrons. The total mass of the protons and neutrons is:<\/li>\n<\/ol>\n\n\n\n

    Mass of protons=11\u00d71.007276470\u2009u=11.08004117\u2009u\\text{Mass of protons} = 11 \\times 1.007276470 \\, \\text{u} = 11.08004117 \\, \\text{u}Mass of protons=11\u00d71.007276470u=11.08004117uMass of neutrons=11\u00d71.008664904\u2009u=11.09531494\u2009u\\text{Mass of neutrons} = 11 \\times 1.008664904 \\, \\text{u} = 11.09531494 \\, \\text{u}Mass of neutrons=11\u00d71.008664904u=11.09531494uTotal mass of nucleons=11.08004117\u2009u+11.09531494\u2009u=22.17535611\u2009u\\text{Total mass of nucleons} = 11.08004117 \\, \\text{u} + 11.09531494 \\, \\text{u} = 22.17535611 \\, \\text{u}Total mass of nucleons=11.08004117u+11.09531494u=22.17535611u<\/p>\n\n\n\n

      \n
    1. Mass defect<\/strong>:
      The mass defect is the difference between the mass of the nucleus and the total mass of its nucleons:<\/li>\n<\/ol>\n\n\n\n

      Mass defect=22.17535611\u2009u\u221221.994437\u2009u=0.18091911\u2009u\\text{Mass defect} = 22.17535611 \\, \\text{u} – 21.994437 \\, \\text{u} = 0.18091911 \\, \\text{u}Mass defect=22.17535611u\u221221.994437u=0.18091911u<\/p>\n\n\n\n

        \n
      1. Binding energy<\/strong>:
        Now, we can convert the mass defect into energy using E=mc2E = mc^2E=mc2. Since 1 atomic mass unit (u) corresponds to 931.5 MeV, the binding energy of the nucleus is:<\/li>\n<\/ol>\n\n\n\n

        Binding energy=0.18091911\u2009u\u00d7931.5\u2009MeV\/u=168.23\u2009MeV\\text{Binding energy} = 0.18091911 \\, \\text{u} \\times 931.5 \\, \\text{MeV\/u} = 168.23 \\, \\text{MeV}Binding energy=0.18091911u\u00d7931.5MeV\/u=168.23MeV<\/p>\n\n\n\n

          \n
        1. Binding energy per nucleon<\/strong>:
          The number of nucleons in Na-22 is 22 (11 protons + 11 neutrons). Therefore, the binding energy per nucleon is:<\/li>\n<\/ol>\n\n\n\n

          Binding energy per nucleon=168.23\u2009MeV22=7.64\u2009MeV\\text{Binding energy per nucleon} = \\frac{168.23 \\, \\text{MeV}}{22} = 7.64 \\, \\text{MeV}Binding energy per nucleon=22168.23MeV\u200b=7.64MeV<\/p>\n\n\n\n

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

          Beta-plus decay involves the conversion of a proton into a neutron, emitting a positron and a neutrino. The reaction is:Na-22\u2009(11,22)\u2192Ne-22\u2009(10,22)+e++\u03bde\\text{Na-22} \\, (11,22) \\rightarrow \\text{Ne-22} \\, (10,22) + e^+ + \\nu_eNa-22(11,22)\u2192Ne-22(10,22)+e++\u03bde\u200b<\/p>\n\n\n\n

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

            \n
          • Na-22 is the sodium-22 nucleus (Z=11),<\/li>\n\n\n\n
          • Ne-22 is the neon-22 nucleus (Z=10),<\/li>\n\n\n\n
          • e+e^+e+ is the emitted positron, and<\/li>\n\n\n\n
          • \u03bde\\nu_e\u03bde\u200b is the neutrino.<\/li>\n<\/ul>\n\n\n\n

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

            The Q-value for a nuclear reaction is the difference in mass between the reactants and products, converted into energy. The general formula is:Q=(Mass of Na-22\u2212Mass of Ne-22\u2212Mass of positron)\u00d7c2Q = (\\text{Mass of Na-22} – \\text{Mass of Ne-22} – \\text{Mass of positron}) \\times c^2Q=(Mass of Na-22\u2212Mass of Ne-22\u2212Mass of positron)\u00d7c2<\/p>\n\n\n\n

              \n
            1. Mass of Na-22<\/strong> = 21.994437 u<\/li>\n\n\n\n
            2. Mass of Ne-22<\/strong> = 21.991386 u<\/li>\n\n\n\n
            3. Mass of positron<\/strong> = 0.000548580 u<\/li>\n<\/ol>\n\n\n\n

              Now, calculate the mass defect:Mass defect=21.994437\u2009u\u2212(21.991386\u2009u+0.000548580\u2009u)=21.994437\u2009u\u221221.991934580\u2009u=0.00250242\u2009u\\text{Mass defect} = 21.994437 \\, \\text{u} – (21.991386 \\, \\text{u} + 0.000548580 \\, \\text{u}) = 21.994437 \\, \\text{u} – 21.991934580 \\, \\text{u} = 0.00250242 \\, \\text{u}Mass defect=21.994437u\u2212(21.991386u+0.000548580u)=21.994437u\u221221.991934580u=0.00250242u<\/p>\n\n\n\n

              Convert this mass defect to energy using 1\u2009u=931.5\u2009MeV1 \\, \\text{u} = 931.5 \\, \\text{MeV}1u=931.5MeV:Q=0.00250242\u2009u\u00d7931.5\u2009MeV\/u=2.33\u2009MeVQ = 0.00250242 \\, \\text{u} \\times 931.5 \\, \\text{MeV\/u} = 2.33 \\, \\text{MeV}Q=0.00250242u\u00d7931.5MeV\/u=2.33MeV<\/p>\n\n\n\n

              Thus, the Q-value for the beta-plus decay is 2.33\u2009MeV2.33 \\, \\text{MeV}2.33MeV.<\/p>\n\n\n\n

              (d) Decay Rate and Mass of Sodium-22<\/h3>\n\n\n\n

              The decay rate (\u03bb\\lambda\u03bb) is related to the half-life (t1\/2t_{1\/2}t1\/2\u200b) by the formula:\u03bb=ln\u2061(2)t1\/2\\lambda = \\frac{\\ln(2)}{t_{1\/2}}\u03bb=t1\/2\u200bln(2)\u200b<\/p>\n\n\n\n

              Given the half-life of Na-22 is 2.603 years, we first convert this into seconds:2.603\u2009years=2.603\u00d7365.25\u00d724\u00d760\u00d760\u2009seconds=8.213\u00d7107\u2009seconds2.603 \\, \\text{years} = 2.603 \\times 365.25 \\times 24 \\times 60 \\times 60 \\, \\text{seconds} = 8.213 \\times 10^7 \\, \\text{seconds}2.603years=2.603\u00d7365.25\u00d724\u00d760\u00d760seconds=8.213\u00d7107seconds<\/p>\n\n\n\n

              Thus:\u03bb=ln\u2061(2)8.213\u00d7107\u2009s=8.43\u00d710\u22129\u2009s\u22121\\lambda = \\frac{\\ln(2)}{8.213 \\times 10^7 \\, \\text{s}} = 8.43 \\times 10^{-9} \\, \\text{s}^{-1}\u03bb=8.213\u00d7107sln(2)\u200b=8.43\u00d710\u22129s\u22121<\/p>\n\n\n\n

              Now, the decay rate (RRR) is related to the mass of the sample by:R=\u03bb\u00d7NR = \\lambda \\times NR=\u03bb\u00d7N<\/p>\n\n\n\n

              Where NNN is the number of atoms in the sample, and we can relate this to the mass of the sample mmm using the molar mass of Na-22 (22.994 u). The number of atoms in the sample is:N=mmolar mass\u00d7NAN = \\frac{m}{\\text{molar mass}} \\times N_AN=molar massm\u200b\u00d7NA\u200b<\/p>\n\n\n\n

              Where NA=6.022\u00d71023\u2009atoms\/molN_A = 6.022 \\times 10^{23} \\, \\text{atoms\/mol}NA\u200b=6.022\u00d71023atoms\/mol. Solving for mmm:127500=8.43\u00d710\u22129\u00d7m22.994\u00d76.022\u00d71023127500 = 8.43 \\times 10^{-9} \\times \\frac{m}{22.994} \\times 6.022 \\times 10^{23}127500=8.43\u00d710\u22129\u00d722.994m\u200b\u00d76.022\u00d71023<\/p>\n\n\n\n

              Solving for mmm, we get:m=3.12\u00d710\u22123\u2009kg=3.12\u2009gm = 3.12 \\times 10^{-3} \\, \\text{kg} = 3.12 \\, \\text{g}m=3.12\u00d710\u22123kg=3.12g<\/p>\n\n\n\n

              (e) Time for 99.9% Decay<\/h3>\n\n\n\n

              The formula for the fraction of a substance remaining after time ttt is:N(t)N0=e\u2212\u03bbt\\frac{N(t)}{N_0} = e^{-\\lambda t}N0\u200bN(t)\u200b=e\u2212\u03bbt<\/p>\n\n\n\n

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

              Taking the natural logarithm of both sides:ln\u2061(0.001)=\u2212\u03bbt\\ln(0.001) = -\\lambda tln(0.001)=\u2212\u03bbt<\/p>\n\n\n\n

              Solving for ttt:t=ln\u2061(0.001)\u2212\u03bb=\u22126.907\u22128.43\u00d710\u22129=8.2\u00d7108\u2009secondst = \\frac{\\ln(0.001)}{-\\lambda} = \\frac{-6.907}{-8.43 \\times 10^{-9}} = 8.2 \\times 10^8 \\, \\text{seconds}t=\u2212\u03bbln(0.001)\u200b=\u22128.43\u00d710\u22129\u22126.907\u200b=8.2\u00d7108seconds<\/p>\n\n\n\n

              Converting this into years:t=8.2\u00d710860\u00d760\u00d724\u00d7365.25=26\u2009yearst = \\frac{8.2 \\times 10^8}{60 \\times 60 \\times 24 \\times 365.25} = 26 \\, \\text{years}t=60\u00d760\u00d724\u00d7365.258.2\u00d7108\u200b=26years<\/p>\n\n\n\n

              Thus, it will take approximately 26 years<\/strong> for 99.9% of the sodium-22 to decay.<\/p>\n\n\n\n

              Summary of Answers:<\/h3>\n\n\n\n
                \n
              • (a) Binding energy per nucleon: 7.64\u2009MeV7.64 \\, \\text{MeV}7.64MeV<\/li>\n\n\n\n
              • (b) Decay reaction: Na-22\u2192Ne-22+e++\u03bde\\text{Na-22} \\rightarrow \\text{Ne-22} + e^+ + \\nu_eNa-22\u2192Ne-22+e++\u03bde\u200b<\/li>\n\n\n\n
              • (c) Q-value: 2.33\u2009MeV2.33 \\, \\text{MeV}2.33MeV<\/li>\n\n\n\n
              • (d) Mass of sample: 3.12\u2009g3.12 \\, \\text{g}3.12g<\/li>\n\n\n\n
              • (e) Time for 99.9% decay: 26\u2009years26 \\, \\text{years}26years<\/li>\n<\/ul>\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-259968","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259968","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=259968"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259968\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=259968"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=259968"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=259968"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}