Question

Stationary nucleus ${}^{238}U$ decays by a emission generating a total kinetic energy T:
${}_{92}^{238}U{\to }_{90}^{234}Th{+}_2^4\alpha$
What is the kinetic energy of the $\alpha -particle$?

• A. Slightly less than $T/2$
• B. $T/2$
• C. Slightly less than $T$
• D. Slightly greater than $T$

Answer 1

The correct option is C Slightly less than $T$

$E=\frac{p^2}{2m}$
$p=\sqrt{2mE}$

Momentum Conservation:
$\sqrt{2m_{\alpha }{E}_{\alpha }}=\sqrt{2m_{Th}{E}_{Th}}$ ------------(1)

Total Kinetic energy =$T$
$T=E_{Th}+E_{\alpha }$ ------------(2)

Substituting:
$\sqrt{2m_{\alpha }{E}_{\alpha }}=\sqrt{2m_{Th}(T-E_{\alpha })}$

$m_{\alpha }{E}_{\alpha }=m_{Th}(T-E_{\alpha })$

$E\left(\alpha \right)=\frac{m_{Th}}{m_{Th}+m_{\alpha }}T$

$E\left(\alpha \right)=\frac{234}{238}T=0.983T$

Slightly less than $T$.