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Question

A neutron collides head-on with a stationary hydrogen atom in ground state. Which of the following statement(s) is / are correct?

(Assume that mass of the neutron and mass of the hydrogen atom are equal)

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Solution

The correct option is **D** If the kinetic energy of the neutron is more than 20.4 eV, collision must be inelastic.

The collision will be inelastic if a part of the initial kinetic energy of the neutron is used to excite the atom.

Let ΔE be used to excite the atom, v be the speed of the neutron before collision, v1 and v2 be the speed of the neutron and the speed of the hydrogen atom after collision.

By conservation of momentum,

mv=mv1+mv2−−−(1)

By conservation of energy,

12mv2=12mv21+12mv22+ΔE−−−(2)

From (1), v2=v21+v22+2v1v2

From (2), v2=v21+v22+2ΔEm

⇒2v1v2=2ΔEm−−−(3)

We know that,

(v1−v2)2=(v1+v2)2−4v1v2

Substituting (1) and (3) in above equation,

(v1−v2)2=v2−4ΔEm

As, (v1−v2) has to be real,

⇒v2>4ΔEm

⇒12mv2>2ΔE

The minimum energy required to excite a hydrogen atom from ground state is 10.2 eV

Thus, the minimum kinetic energy of the neutron before collision required for inelastic collision should be 20.4 eV

Hence, option (D) is correct.

The collision will be inelastic if a part of the initial kinetic energy of the neutron is used to excite the atom.

Let ΔE be used to excite the atom, v be the speed of the neutron before collision, v1 and v2 be the speed of the neutron and the speed of the hydrogen atom after collision.

By conservation of momentum,

mv=mv1+mv2−−−(1)

By conservation of energy,

12mv2=12mv21+12mv22+ΔE−−−(2)

From (1), v2=v21+v22+2v1v2

From (2), v2=v21+v22+2ΔEm

⇒2v1v2=2ΔEm−−−(3)

We know that,

(v1−v2)2=(v1+v2)2−4v1v2

Substituting (1) and (3) in above equation,

(v1−v2)2=v2−4ΔEm

As, (v1−v2) has to be real,

⇒v2>4ΔEm

⇒12mv2>2ΔE

The minimum energy required to excite a hydrogen atom from ground state is 10.2 eV

Thus, the minimum kinetic energy of the neutron before collision required for inelastic collision should be 20.4 eV

Hence, option (D) is correct.

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