A neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest. assume: the mass of the atomic nucleus is about 14.9 the mass of the neutron. what fraction of the neutron's kinetic energy is transferred to the atomic nucleus?

Here, m, M are the masses of the neutron and atom, [tex]v_{i}[/tex] is the initial velocity of the neutron,[tex]v_{f}[/tex] is the final velocity of the neutron and [tex]V_{f}[/tex] is the final velocity of the atom.

Use the conservation of the kinetic energy.

[tex]\frac{1}{2}mv_{i}^{2}=\frac{1}{2}mv_{f}^{2}+\frac{1}{2}MV_{f}^{2}[/tex]

[tex]v_{i}^{2}-v_{f}^{2}=\frac{M}{m}V_{f}^{2}[/tex]

Put

[tex]\frac{M}{m}=14.9[/tex].

[tex]v_{i}^{2}-v_{f}^{2}=14.9V_{f}^{2}[/tex] ...... (2)

Divide (2) by (1).

[tex]\frac{v_{i}^{2}-v_{f}^{2}}{v_{i}-v_{f}}=\frac{14.9V_{f}^{2}}{14.9V_{f}}[/tex]

[tex]\frac{v_{i}^{2}-v_{f}^{2}}{v_{i}-v_{f}}=V_{f}[/tex]

[tex]v_{i}+v_{f}=V_{f}[/tex] ...... (3)

Solve equation (1) and (3).

[tex]v_{i}=7.9V_{f}[/tex]

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

[tex]Fraction\ transferred\ to\ atom =\frac{\frac{1}{2}MV_{f}^{2}}{\frac{1}{2}mv_{i}^{2}}[/tex]

Put

[tex]\frac{M}{m}=14.9[/tex]

[tex]Fraction\ transferred\ to\ atom=\frac{14.9}{7.95}[/tex]

[tex]Fraction\ transferred\ to\ atom=1.87[/tex]

Therefore, the fraction of the neutron's kinetic energy transferred to the atomic nucleus is 1.87.

priyankatutor priyankatutor · Answer 2 · 2021-10-30T12:47:57+02:00

In elastic head-on collision, the energy of the system and total momentum is conserved. The fraction of the neutron's kinetic energy transferred to the atomic nucleus is 1.87.

In elastic head-on collision, the energy of the system and total momentum is conserved.

For the neutron,

m - mass

[tex]\bold { v_i}[/tex] - initial velocity
[tex]\bold { v_f}[/tex] - final velocity

For the atom ,

M- mass
[tex]\bold { V_i}[/tex] - initial velocity
[tex]\bold { V_f}[/tex] - final velocity

Conserved momentum on the head-on collision

[tex]\bold {mv_i = mv_f + MV_f}\\\\\bold {v_i - v_f =MmV_f}\\\\\bold { v_i - v_f = 14.9Vf}[/tex].........(I)

The kinetic energy

[tex]\bold { v_i^2 - v_f^2 = 14.9v_f^2}[/tex].............(II)

From equation (I) and (II)

[tex]\bold{\frac{ v_i - v_f}{v_i ^2- v_f^2} = \frac{12v_f^2}{12v_f} }\\\\\bold{ v_i + v_f = v_f }[/tex]...........(III)

Solve equation (I) and (III) for [tex]\bold { v_i}[/tex]

[tex]\bold{v_i = 7.9v_f }[/tex]

Now, the fraction of the neutron's kinetic energy transferred to the atomic nucleus,

[tex]\rm \bold { \frac{\frac{1}{2} MV_f^2}{\frac{1}{2} mv_i^2} = \frac{14.9}{7.95} = 1.87 }[/tex]

Hence, we can conclude that the fraction of the neutron's kinetic energy transferred to the atomic nucleus is 1.87.

To know more about kinetic energy, refer to the link:

https://brainly.com/question/12669551