BITSAT 2015 Paper

Let speed of neutron before collision = V
Speed of neutron after collision = V1
Speed of proton or hydrogen atom after collision = V2
Energy of excitation = DE
From the law of conservation of linear momentum,
mv = mv1+mv2 ...(1)
And for law of conservation of energy,

1

2

mv2 =

1

2

mv12+

1

2

mv22 ... (2)
From squaring eq. (i), we get
v2 = v12+v22+2v1v2 ... (3)
From squaring eq. (ii), we get
v2 = v12+v22 +

2ΔE

m

∴ (v1−v2)2 = (v1+v2)2 - 4v1v2
= v2−

4ΔE

m

As, v−1−v2 must be real, v2−

4ΔE

m

≥ 0
⇒

1

2

mv2 ≥ 2ΔE
The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is 10.2 eV. Therefore, the maximum kinetic energy needed is

1

2

mvmin2 = 2 × 10.2 = 20.4 eV