Initially the nucleus of radium 226 is at rest. It decays due to which and `alpha` 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)=4.002"amu",m_("Rn")=22.017"amu"]`
What is the linear momentum of the `alpha` -particle?

Options

A. `2.01xx10^(19)kgms^(-1)`
B. `1.01xx10^(-19)kgms^(-1)`
C. `4.01xx10^(-19)kgms^(-1)`
D. `8.01xx10^(-19)kgms^(-1)`

Answer

B

Explanation

`P=mv` &
`(P^(2))/(2m)=1/2mV^(2)`
`E_(alpha)=(P^(2))/(2m)`
`E_("radon")=(P^(2))/(2M)`
Also
`(P^(2))/(2m)+(P^(2))/(2M)=/_\EimpliesP=sqrt(2/_\E(mM)/(m+M))`
`m=4.002"mu", M=222.017 "mu"`
`m=6.64xx10^(-27)kg`