An `alpha`-particlee of mass m suffers one dimensional elastic collision with a nucleus of unknown mass. After the collision the `alpha`-particle is scattered directly backwards losing 75% of its kinetic energy. Then, the mass of the nucleus is

Options

A. m
B. 2m
C. 3m
D. `3/2` m

Answer

D

Explanation

In the reaction, we have to conserve the mass number and the atomic number. Thus, we have
`""_(7)^(14)N+""_(b)^(a)X to ""_(6)^(14)C+""_(1)^(1)H`
so, `14+a=14+1`
hence, `a=1 and 7+b=6+1`
Hence, b=0
So, we have, `""_(b)^(a)X=""_(0)^(1)X=""_(0)^(1)n`.