Question

$\alpha$-particle is emitted from a ${}_{88}^{226}Re$ nucleus. If the energy of $\alpha$-particle is 4.662 MeV, then what is the total free energy in this decay?

Answer 1

${}_{88}R{a}^{226}$

$Z=88$

$A=226$

$E_{\alpha }=4.662MeV$

$E_{\alpha }=\frac{A-4}{A}Q$

$Q=\left(\frac{A-4}{A}\right)E_{\alpha }$

$Q=\left(\frac{226}{226-4}\right)4.662MeV$

$Q=\frac{226}{222}\times 4.662MeV$

$Q=4.746MeV$