After absorbing a slowly moving neutrons of mass `m_(N)` (momentum ~0) a nucleus of mass M breaks into two nuclie of mass `m_(1) and 5m_(1)(6m_(1)=M+m_(N))`, respectively . If the de-Broglie wavelength of the nucleus with mass `m_(1) is lambda`, then de Broglie wavelength of the other nucleus will be

Options

A. `25lambda`
B. `5lambda`
C. `lambda//5`
D. `lambda`

Answer

d

Explanation

Initial momentum of nucleus of mass `M,P_(i)=0`. Let `P_(1)` and `P_(2)` be momentum of broken nuclei of masses `m_(1)` and `5m_(1)` respectively.
According to law of conservation of momentum
`P_(i)=P_(1)+P_(2) :. 0=P_(1)+P_(2)` or `P_(1)=-P_(2)`.
Hence `lambda_(1)=h/(P_(1))`
and `lambda_(2)=h/(P_(2))=h/(P_(1))` (in magnitude)
`:. lambda_(1)=lambda_(2)=lambda`