A nuclider X, initally at rest, undergoes `alpha` -decay according to the equation `92^(X^(Ato))z ^(Y^(238))+ sigma.`
The `alpha` -particle produced in the above process is found to move in a circular track of radiuc `0.11 ` m in a uniform magnetic field of 3T. Find the binding energy per nucleon (in Mev) of the daughter nuclide Y. Given that `(m (y) =228.03u, m(_(2)He^(4)) =4.003u, m (_(0)n^(1)) =1.009 u , m (_(1)H^(1)) =1 amu = 931.5 MeV//c^(2) =1.66 xx10^(-27)kg`

Options

A. `5.89` meV
B. `6.89` MeV
C. `8.89` MeV
D. `7.89` MeV

Answer

D

Explanation

`(B.E) _(y) =(90xx1.008+138xx1.009 -228.03)u=1.932 u`
`=1.932 xx931 .5 MeV`
Hence, Binding energy is per Nucleon is
`= (1.932xx931.5)/(228)=7.89 MeV`