Question

Initially, the nucleus of $\text{Radium}-226$ is at rest. It decays by emitting an $\alpha -$ particle, and thus the nucleus of $\text{Radon}$ is created. The released energy during the decay is $4.87\text{MeV}$, which appears as the kinetic energy of the two resulted particles. The kinetic energy of $\alpha -$ particle & $\text{Radon}$ nucleus are respectively.
$[m_{\alpha }=4.002u,m_{\text{Rn}}=222.017u]$

• A. $3.08\text{MeV},0.09\text{MeV}$
• B. $0.09\text{MeV},4.78\text{MeV}$
• C. $3.68\text{MeV},1.09\text{MeV}$
• D. $4.78\text{MeV},0.09\text{MeV}$

Answer 1

The correct option is D $4.78\text{MeV},0.09\text{MeV}$

Given:
$Q=4.87\text{MeV}$

The given decay can be written as,

${\text{Ra}}^{226}\to {\text{Rn}}^{222}+\alpha$

$m_{\alpha }=4.002u,m_{\text{Rn}}=222.017u$

So, kinetic energy of $\alpha -$ particle is given by,

$K.E_{\alpha }=\frac{m_{\text{Rn}}}{m_{\alpha }+m_{\text{Rn}}}Q=\frac{222.017}{226.019}\left(4.87\text{MeV}\right)$

$\Rightarrow K.E_{\alpha }=4.78\text{MeV}$

$\therefore K.E_{\text{Rn}}=4.87-4.78=0.09\text{MeV}$

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