The isotope ${ }_5^{12} B$ having a mass 12.014 u undergoes $\beta$ - decay to ${ }_6^{12} C .{ }_6^{12} C$ has an excited state of the nucleus $\left({ }_6^{12} C{ }^*\right)$ at $4.041 \mathrm{MeV}$ above its ground state. If ${ }_5^{12} B$ decays to ${ }_6^{12} C^*$, the maximum…

Question

The isotope ${ }_5^{12} B$ having a mass 12.014 u undergoes $\beta$ - decay to ${ }_6^{12} C .{ }_6^{12} C$ has an excited state of the nucleus $\left({ }_6^{12} C{ }^*\right)$ at $4.041 \mathrm{MeV}$ above its ground state. If ${ }_5^{12} B$ decays to ${ }_6^{12} C^*$, the maximum kinetic energy of the $\beta$-particle in units of $\mathrm{MeV}$ is ($1 u=931.5 M e \frac{V}{c^2}$, where c is the speed of light in vacuum).

Aicharya · Accepted Answer

B &#8594; 5 12 &#160; C 6 12 + &#160; e - 1 0 + &#160; v - Mass defect =(12.014 - 12) u &#8756; Released energy = 13.041 MeV Energy used for excitation of C 6 12 = 4.041 &#160; M e V &#8756; Energy converted to KE of electron = 13.041 - 4.041 = 9 &#160; M e V The β particle will have maximum K.E when antineutrino will have minimum K.E i.e 0. Therefore maximum K.E of the β particle is 9 MeV ;

The isotope \({ }_5^{12} B\) having a mass 12.014 u undergoes \(\beta\) - decay to \({ }_6^{12} C .{…

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