Question 205: Centre of Mass and Rotational Motion

Question

A neutron moving with a speed ‘v’ makes a head on collision with a stationary
hydrogen atom in ground state. The minimum kinetic energy of the neutron for
which inelastic collision will take place is :

Accepted Answer

Let, velocity offer collision = v
1
$$ 	herefore $$   From conservation of momentum,
mv = (m + m) v
1
$$ \Rightarrow $$    v
1
= $${v \over 2}$$
$$ 	herefore $$   Loss in kinetic energy
= $${1 \over 2}$$ mv
2
$$-$$ $${1 \over 2}$$ (2m) $$ 	imes $$ $${\left( {{v \over 2}} ight)^2}$$
= $${1 \over 4}\,$$mv
2
lost kinetic energy is used by the electron to jump from first  orbit to second orbit.
$$ 	herefore $$   $${1 \over 4}$$mv
2
= (13.6 $$-$$ 3.4) eV = 10.2 eV
$$ \Rightarrow $$  $${1 \over 2}$$mv
2
= 20.4 eV

Answer

10.2 eV

Answer

16.8 eV

Answer

12.1 eV

Answer

20.4 eV

Centre of Mass and Rotational Motion

A neutron moving with a speed ‘v’ makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which inelastic collision will take place is :

Choose the correct answer: