13.14 The nucleus ${}_{10}{}^{23}\mathit{N}\mathit{e}$ decays by ${\mathit{\beta }}^{-}$ emission. Write down the $\mathit{\beta }$ -decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m ( ${}_{10}{}^{23}\mathit{N}\mathit{e})$ = 22.994466 u

m ( ${}_{11}{}^{23}\mathit{N}\mathit{a})$ = 22.989770 u.

<p><strong>13.14&nbsp;</strong>In&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>β</mi></mrow></mrow><mrow><mrow><mo>-</mo></mrow></mrow></msup></math></span>&nbsp;emission, the number of protons increase by 1 and one electron and an antineutrino are emitted from the parent nucleus.</p><p><span title="Click to copy mathml"><math><msup><mrow><mrow><mi>β</mi></mrow></mrow><mrow><mrow><mo>-</mo></mrow></mrow></msup></math></span>&nbsp;emission of the nucleus&nbsp;<span title="Click to copy mathml"><math><mmultiscripts><mrow><mrow><mi>N</mi><mi>e</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></mmultiscripts></math></span>&nbsp;:</p><p><span title="Click to copy mathml"><math><mmultiscripts><mrow><mrow><mi>N</mi><mi>e</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></mmultiscripts></math></span>&nbsp;<span title="Click to copy mathml"><math><mo>→</mo><mmultiscripts><mrow><mrow><mi>N</mi><mi>a</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>11</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></mmultiscripts></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>e</mi></mrow></mrow><mrow><mrow><mo>-</mo></mrow></mrow></msup></math></span>&nbsp;+&nbsp;<span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><mi>ν</mi></mrow></mrow><mo>?</mo></mover></math></span>&nbsp;+ Q</p><p>It is given that:</p><p>Atomic mass m (&nbsp;<span title="Click to copy mathml"><math><mmultiscripts><mrow><mrow><mi>N</mi><mi>e</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></mmultiscripts><mo>)</mo></math></span>&nbsp;= 22.994466 u</p><p>Atomic mass m (&nbsp;<span title="Click to copy mathml"><math><mmultiscripts><mrow><mrow><mi>N</mi><mi>a</mi><mo>)</mo></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>11</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></mmultiscripts></math></span>&nbsp;= 22.989770 u</p><p>Mass of an electron,&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></math></span>&nbsp;= 0.000548 u</p><p>Q value of the given reaction is given as Q =&nbsp;<span title="Click to copy mathml"><math><mfenced open="[" close="]" separators="|"><mrow><mrow><mi>m</mi><mi></mi><mfenced separators="|"><mrow><mrow><mmultiscripts><mrow><mrow><mi>N</mi><mi>e</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></mmultiscripts></mrow></mrow></mfenced><mo>-</mo><mi></mi><mo>{</mo><mi>m</mi><mi></mi><mo>(</mo><mmultiscripts><mrow><mrow><mi>N</mi><mi>a</mi><mo>)</mo></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>11</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></mmultiscripts><mo>+</mo><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub><mo>}</mo><mi></mi></mrow></mrow></mfenced><msup><mrow><mrow><mi>c</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span></p><p>There are 10 electrons in&nbsp;<span title="Click to copy mathml"><math><mmultiscripts><mrow><mrow><mi>N</mi><mi>e</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></mmultiscripts></math></span>&nbsp;and 11 electrons in&nbsp;<span title="Click to copy mathml"><math><mmultiscripts><mrow><mrow><mi>N</mi><mi>a</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>11</mn></mrow></mrow><mrow><mrow><mn>23</mn></mrow></mrow></mmultiscripts></math></span>&nbsp;. Hence, the mass of the electron is cancelled in the Q-value equation.</p><p>Therefore Q = {22.994466 - 22.989770}&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>c</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span>&nbsp;= 4.696&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup><msup><mrow><mrow><mi>c</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span>&nbsp;u</p><p>But 1 u = 931.5 MeV/&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>c</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span></p><p>Q = 4.374 MeV</p><p>The daughter nucleus is too heavy as compared to&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>e</mi></mrow></mrow><mrow><mrow><mo>-</mo></mrow></mrow></msup></math></span>&nbsp;and&nbsp;<span title="Click to copy mathml"><math><mover accent="true"><mrow><mrow><mi>ν</mi></mrow></mrow><mo>?</mo></mover></math></span>&nbsp;. Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e. 4.374 MeV.</p>

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