Question

Initially the nucleus of radium – 226 is at rest. It decays due to which and a particle and the nucleus of radon are created. The released energy during the decay is 4.87 Mev, which appears as the kinetic energy of the two resulted particles. $[m_{\alpha }=4.002amu,m_{Rn}=222.017amu]$

What is the linear momentum of the a – particle?

• A. $2.01\times {10}^{-19}kgm{s}^{-1}$
• B. $1.01\times {10}^{-19}kgm{s}^{-1}$
• C. $4.01\times {10}^{-19}kgm{s}^{-1}$
• D. $8.01\times {10}^{-19}kgm{s}^{-1}$

Answer 1

Answer: (C), (D)

The correct options are
C

$4.01\times {10}^{-19}kgm{s}^{-1}$

D

$8.01\times {10}^{-19}kgm{s}^{-1}$

$P=mv$ and
$\frac{P^2}{2m}=\frac{1}{2}m{V}^2{E}_a=\frac{P^2}{2m}{E}_{radon}=\frac{P^2}{2M}$
Also
$\frac{P^2}{2m}+\frac{P^2}{2M}=\Delta E\Rightarrow P=\sqrt{2\Delta E\frac{mM}{m+M}}m=4.002mu,M=222.017mum=6.64\times {10}^{-27}kg$