Question

The isotope ${}_5^{12}\text{B}$ having a mass $12.014\text{u}$ undergoes $\beta -$ decay to ${}_6^{12}\text{C}.{}_6^{12}\text{C}$ has an excited state of the nucleus ${(}_6^{12}{\text{C}}^{\ast })$ at $4.041\text{MeV}$ above its ground state. If ${}_5^{12}\text{B}$ decay to ${}_6^{12}{\text{C}}^{\ast },$ the maximum kinetic energy of the $\beta -$ particle in units of $\text{MeV}$ nearly is (integer only).

Take : $m{(}_6^{12}\text{C})=12.0\text{u}(1u=931.5\text{MeV}/c^2,\text{Where c is the speed of light in vacuum)}$

• A. 9.0
• B. 9.00
• C. 9

Answer 1

Answer: (A), (B), (C)

Given, $m{(}_5^{12}\text{B})=12.014\text{u}m{(}_6^{12}\text{C})=12.0\text{u}{E}_{excitation}=4.041\text{MeV}$

${}_5^{12}\text{B}\to {}_6^{12}\text{C}+{\beta }^{-}+{\gamma }^{-}+Q$

$Q=\Delta m{c}^2=\left[m{(}_5^{12}\text{B}\right)-m{(}_6^{12}{\text{C}}^{\ast }\left)\right]c^2$

$=[12.014-12.0]c^2$

$=0.014\times 931.5=13.041\text{MeV}$

Now, $K{E}_{max}=Q-E_{excitation}$

$K{E}_{max}=13.041-4.041=9\text{MeV}$