Question

A stationary U238 nucleus emits an a-particle ( He*) with velocity 1.4 x 107 m/s and kinetic energy 4.1 Mev. Calculate the velocity of the residual nucleus.​

Answer 1

Explanation:

mass of alpha particle m

a

=4m, where m is mass of a proton.

mass of Uranium nucleus is m

u

=238m after emission the residue will have mass m

r

=238−4=234m

We know KE=momentum

2

/2m

so for residue nucleus, it should be like this E=p

r

2

/2m

r

or momentum of residue nucleus is p

r

=

2

2m

r

E

=

2

468mE

But during emission the alpha particle should get same momentum in opposite direction to make net momentum as zero

before and after emission so it should also be p

a

=

2

468mE

If kinetic energy of alpha particle is E

a

then E

a

=p

a

2

/2m

a

=

2×4m

468mE

=58.5E

So net energy released in the emission is the sum of above two i.e. 59.5E