Question

The K.E. of the emitted $\alpha -particle$ in the decay of ${}_{88}^{226}Ra$ (approximately)

• A. 2.3 MeV
• B. 4.85 MeV
• C. 9.7 MeV
• D. 14 MeV

Answer 1

The correct option is B 4.85 MeV

$\begin{array}{c}{}_{88}^{226}Ra⟶{}_2^{4}H{e}^{2+}+{}_{86}^{222}Rn\end{array}$

$226.02540\mu 4.00260\mu 222.01750\mu$

$\begin{array}{cc}\Delta m& =0.0053\\ \text{Energy}& =0.0053\times 931.5MeV\\ & =4.94\\ K_1& =\frac{m_2}{m_1+m_2}\times K\\ & =\frac{4}{222+4}\times 4.94\\ & =4.8495\end{array}$