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  •  beta-decays to an excited state of <img alt="{ }^{12} \mathrm{C}" src="/latex-image/?%7B%20%7D

{ }^{12} \mathrm{~N} beta-decays to an excited state of { }^{12} \mathrm{C}, which subsequently decays to the ground state with the emission of a 4.43 \mathrm{MeV} gamma ray. The maximum kinetic energy of the emitted beta -particle.

Given: mass of { }^{12} \mathrm{~N}=12.018613 \mathrm{u}, mass of e=0.000549 u.
 

Option: 1

11.9 \mathrm{MeV}


Option: 2

20.4 \mathrm{MeV}


Option: 3

12.80 \mathrm{MeV}


Option: 4

15.5 \mathrm{MeV}


Answers (1)

best_answer

To determine the Q value for this decay, we first find the mass of the product nucleus { }^{12} \mathrm{C}  in its excited state. In the ground state,{ }^{12} \mathrm{C} has a mass of 12.000000 \mathrm{u}, so its mass in the excited state is

12.000000 \mathrm{u}+\frac{4.43 \mathrm{MeV}}{931.5 \mathrm{MeV} / \mathrm{u}}=12.004756 \mathrm{u}
In this decay, a proton is converted to a neutron, so it must be an example of positron decay. The Q value is thus given by

\begin{gathered} \mathrm{Q}=\left[\mathrm{m}\left({ }_7^{12} N\right)-\mathrm{m}\left({ }_6^{12} C^*\right)-\mathrm{m}(\mathrm{e})\right] \mathrm{c}^2 \\ =\left[\mathrm{m}\left({ }^{12} \mathrm{~N}\right)+7 \mathrm{~m}(\mathrm{e})-7 \mathrm{~m}(\mathrm{e})-\mathrm{m}\left({ }_6^{12} C^*\right)-\mathrm{m}(\mathrm{e})\right] \mathrm{c}^2 \\ =\left[\mathrm{m}^{\prime}\left({ }_7^{12} N\right)-\mathrm{m}^{\prime}\left({ }_6^{12} C^*\right)-2 \mathrm{~m}(\mathrm{e})\right] \mathrm{c}^2 \end{gathered}
where m^{\prime} refers to the mass of the atoms.
Thus the Q value of is given by

\begin{aligned} \mathrm{Q}= & {[12.018613 \mathrm{u}-12.004756 \mathrm{u}-2 \times 0.000549 \mathrm{u}] \times 931.5 \mathrm{MeV} / \mathrm{u} } \\ & =11.89 \mathrm{MeV} \end{aligned}

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Anam Khan

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