Question

A neutron in a reactor makes an elastic head on collision with the nucleus of an atom initially at rest.Assume: The mass of the atomic nucleus isabout 12.5 the mass of the neutron.What fraction of the neutron’s kinetic energy is transferred to the atomic nucleus?

Answer 1

0.274348

Step-by-step explanation

Step 1

in a head-on elastic collision between a small projectile, in this case the neutron, and a much more massive target (atom), the projectile will bounce back with essentially the same speed and the massive target will be given a very small velocity,In elastic head-on collision, the energy of the system and total momentum is conserved

it can be represented by the expression

`\begin{gathered} mv_i=mv_(f+MV_f) \\ where \\ m\text{ is the mass of the neuton } \\ v_i=\text{ initial velocity of the neutron} \\ v_f=\text{ final }velocity \\ M\text{ is the mass of the atom} \\ V_{F\text{ }}is\text{ the final velocity of the atom} \end{gathered}`

then, let's set the equations

`\begin{gathered} mv_i=mv_(f+MV_f) \\ \text{subtract mv}_f\text{ in both sides} \\ mv_(i-)\text{mv}_f=mv_(f+MV_F)-\text{mv}_f \\ mv_(i-)\text{mv}_f=MV_F \\ \text{factorize m } \\ m(v_i-v_f)=MV_F \\ \text{divide both sides by m} \\ (m(v_i-v_f))/(m)=(MV_F)/(m) \\ (v_i-v_f)=(M)/(m)V_F \\ as\text{ the ratio ot the mas iss 12.5} \\ (v_i-v_f)=12.5V_{}\Rightarrow equation\text{ (1)} \end{gathered}`

Step 2

the kinetic energy

`\begin{gathered} (1)/(2)mv^2_i=(1)/(2)mv^2_f+(1)/(2)MV^2 \\ (1)/(2)mv^2_i-(1)/(2)mv^2_f=(1)/(2)MV^2 \\ (1)/(2)m(v^2_i-v^2_(_f))=(1)/(2)MV^2_{} \\ v^2_i-v^2_(_f)=(MV^2)/(m) \\ v^2_i-v^2_(_f)=12.5V^2_{}\Rightarrow equation(2) \end{gathered}`

combinde

a)

`\begin{gathered} (v_i-v_f)=12.5V_{}\Rightarrow equation\text{ (1)} \\ v^2_i-v^2_(_f)=12.5V^2_{}\Rightarrow equation(2) \\ v^2_i-v^2_(_f)=(v_i-v_f)(v_i+v_f) \\ \text{hence} \\ (v_i-v_f)(v_i+v_f) \\ v_1+v_f=V\Rightarrow equation(3) \end{gathered}`

Step 3

solve the equaions

`\begin{gathered} from\text{ (1) and (3)} \\ (v_i-v_f)=12.5V \\ v_f=-12.5V+v_i \\ v_(_f)=V-v_1 \\ -12.5V+v_1=V-v_1 \\ -13.5V=-2v_1 \\ \text{divide both sides by -2} \\ (-13.5V)/(-2)=(-2v_1)/(-2) \\ 6.75V=v_i \\ v_i=6.75V \end{gathered}`

Calculate the fraction of the neutron's kinetic energy transferred to the atomic nucleus.

Let

`\begin{gathered} (M)/(m)=12.5 \\ \text{fraction transfered to the atom} \\ ((1)/(2)MV^2)/((1)/(2)mv^2_i) \\ ((1)/(2)MV^2)/((1)/(2)mv^2_i)=12.5(V^2)/((6.75V)^2)=(12.5)/(45.5625)=0.274348 \\ \end{gathered}`

therefore, the answer is 0.274348

I hope this helps you