A radioactive nucleus initially at rest decays by emitting an electron and neutron at right angle to one another. The momentum of neutron is `3.2xx10^(-24)kgm//s` and the momentum of neutron is `6.4xx10^(-24)kgm//sec` . The direction of the recoiling nucleus with that of the electron motion is:

Options

A. `pi-tan^(-1)(2)`
B. `tan^(-1)(2)`
C. `tan^(-1)(0.5)`
D. `pi/2+tan^(-1)(2)`

Answer

A

Explanation

`P_(e)=3.2xx10^(-23)kgm//s` , `P_(n)=6.4xx10^(-23)kgm//s` .
`P_(R)=sqrt(P_(e)^(2)+P_(n)^(2))=sqrt((3.2)^(2)+(6.4)^(2))xx10^(-23)kgm//s` .
`tantheta=(P_(n)/(P_(e))=(6.4xx10^(-23))/(3.2xx10^(-23))=2` .
According to law of conservation of momentum, the residual nucleus must move in a direction just opposite to that of `vec(P)_(R)` , as shown in figure. hence, direction of the recoiling nucleus with that of the electron motion `phi=pi-theta=pi-tan^(-1)(2)` .