A nucleus X, initially at rest , undergoes alpha dacay according to the equation ,
` _(92)^(A) X rarr _(Z)^(228)Y + alpha `
(a) Find the value of `A` and `Z` in the above process.
(b) The alpha particle produced in the above process is found to move in a circular track of radius `0.11 m` in a uniform magnetic field of `3` Tesla find the energy (in MeV) released during the process and the binding energy of the parent nucleus X
Given that `: m (Y) = 228.03 u, m(_(0)^(1)n) = 1.0029 u. `
`m (_(2)^(4) He) = 4.003 u , m (_(1)^(1) H) = 1.008 u `

Options

A. `2.3`
B. `4.7`
C. 6
D. `7.8`

Answer

D

Explanation

Mass lost during a-decay, `Deltam=(E)/(931)=0.005U`
`:.` Mass of nucleus X,
`m_(x)=m_(y)+m_(alpha)+Deltam=225.038 U`
`:.` mass defect in nucleus X,
`m_(d) = {[92m_(p)+(225-92)m_(n)]-m_(x)}U=1.895 U`
`:. BE` per nucleon in nucleus `X= (m_(d)xx931)/(225)=7.84 MeV`