Question

${}_{94}{\text{Pu}}^{239}$ is undergoing $\alpha -$ decay according to the equation

${}_{94}{\text{Pu}}^{239}\to {}_{92}{\text{U}}^{235}+{}_2{\text{He}}^4$

The energy released in the process is mostly kinetic energy of the $\alpha -$ particle. However, a part of the energy is released as $\gamma -$ rays. What is the speed of the emitted $\alpha -$ particle if the $\gamma -$rays radiated out have energy of $0.90\text{MeV}?$

Given:
Mass of ${}_{94}{\text{Pu}}^{239}=239.05122u$
Mass of ${}_{92}{\text{U}}^{235}=235.04299u$
Mass of${}_2{\text{He}}^4=4.002602u$
$(1u{c}^2=931\text{MeV})$

• A. $1.4\times {10}^7{\text{ms}}^{-1}$
• B. $2.8\times {10}^7{\text{ms}}^{-1}$
• C. $3.5\times {10}^7{\text{ms}}^{-1}$
• D. $6\times {10}^7{\text{ms}}^{-1}$

Answer 1

The correct option is A $1.4\times {10}^7{\text{ms}}^{-1}$

Total energy released during the decay
$Q=\Delta m{c}^2$

$\Rightarrow Q=[m\left({}_{94}{\text{Pu}}^{235}\right)-m({}_{92}{\text{U}}^{235}+{}_2{\text{He}}^4)]c^2$

$\Rightarrow Q=[239.05122u-235.04299u-4.002602u]c^2$

$\Rightarrow Q=0.005628u{c}^2$

$\Rightarrow Q=0.005628\times 931\text{MeV}$

$\Rightarrow Q=5.239668\text{MeV}$

As we know that,

$Q=E_U+E_{\alpha }+E_{\gamma }$

Since, $E_U$ is negligible,

$\therefore Q=E_{\alpha }+E_{\gamma }$

$\Rightarrow E_{\alpha }=5.239668\text{MeV}-0.90\text{MeV}$

$\Rightarrow \frac{1}{2}{m}_{He}{v}^2=4.339668\text{MeV}$

$\Rightarrow v^2=\frac{2\times 4.339668\text{MeV}}{4.002602u}$

$\Rightarrow v^2=\frac{2\times 4.339668}{4.002602}\frac{c^2}{931}$

$\Rightarrow v^2=0.002329\times c^2$

$\Rightarrow v=0.048259\times c$

$\Rightarrow v=1.44\times {10}^7{\text{ms}}^{-1}$

Hence, option $\left(\text{A}\right)$ is correct.