Manipal Entrance Test 2015 Physics Paper

Let v = speed of neutron before collision
v1=speed of neutron after collision
v2= speed of hydrogen atom after collision and
ΔE= energy of the excitation
From the conservation of linear momentum,
mv=mv1+mv2 ...(i)
From the conservation of energy,

1

2

mv2=

1

2

mv12+

1

2

mv22+ΔE ...(ii)
from Eq. (i),
v2=v12+v22+2v1v2
From Eq. (ii),
v2=v12+v22+

2ΔE

m

∴2v1v2=

2ΔE

m

∴(v1−v2)2=(v1+v2)2−4v1v2
=v2−

4ΔE

m

As, v1−v2 must be real
⇒v2−

4ΔE

m

≥0
or

1

2

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

1

2

mvmin2 =2×10.2‌eV =20.4‌eV