Question

A moving neutron collides with a stationary alpha particle. the fraction of the kinetic energy lost by the neutron is

Answer 1

The energy lost by neutron is 16/25.

Let mass of neutron be = m

Mass of the alpha particle to the mass of neutron = 4 times

Therefore, mass of alpha particle = 4m

Since, light particle collides with the massive particle, thus -  

vf = mn - ma / mn + ma.vi

Initial and final kinetic energy of neutron -  

Ki = 1/2mvi² and Kf = 1.2mvf2²

= 1/2 ( mn - ma / mn + ma).vi ²

Kinetic energy lost = Ki - Kf/Ki = 1 - (mn - ma / mn + ma)

= 4mnma/ (mn+ma)

Fraction of energy lost = 4 × 1 × 1 / ( 1+4 ) ²

= 16/25

Thus, the energy lost by neutron is 16/25.

Answer 2

The fraction of the kinetic energy lost by the neutron is $\frac{16}{25}.$

Explanation:

Let Vi be the initial velocity that is velocity of neutron and Vf be the final velocity or velocity After collision.

                                            Vf =   $\frac{(mn-ma)}{(mn+ma)} Vi$

So final and initial velocity of neutron will be given as,

                                                Ki =  $\frac{1}{2} mn Vi^2$

                                                Kf = $\frac{1}{2} mn Vf^2$

So, the final kinetic energy that Alpha particle will have will be the fraction that was lost by neutron during collision.

                                               = $1 - {\frac{(mn-ma)}{(mn+ma)}}^{2}$

                                                 ⇒ $\frac{4mnma}{(mn+ma)}^2$

                                                  ⇒ $\frac{4\times 1 \times 4}{(1+4)}^2$

                                  ⇒ $\frac{16}{25}.$

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