After absobing a slowly moving neutron o...

After absobing a slowly moving neutron of mass `m_(N)` (mometum `~0`), a nucleus of mass M breaks into two nuclei of masses `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

`25 lambda`

`5 lambda`

`(lambda)/(5)`

`lambda`

Text Solution

Verified by Experts

The correct Answer is:

According to Conservation of linear momentum `P=P`. Therefore `lambda'=lambda`

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