A neutron travelling with a velocity v and kinetic energy E collides elastically head on with the nucleus of an atom of mass number A at rest. The fraction of total energy retained by the neutron is

- (A-1 / A+1)
^{2} - (A+1 / A-1)
^{2} - (A-1 / A)
^{2} - (A+1 / A)
^{2}

Hello!

Let say that a neutron with mass 'm_{1}' and initial speed 'u_{1}' strikes the nucleus (which is at rest, so u_{2} = 0) and undergoes a perfectly elastic collision. The final velocity (v_{1}) of the neutron would be given as (from combining the Laws of Conservation of Kinetic energy and Momentum)

In this case, u_{2} = 0, thus

or

taking m_{1} common and canceling it , we get

or

here A is the mass number of the nucleus.

Now, the fraction of kinetic energy retained would be

or

So, **the answer would be option (1)**

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