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. `2.5`
B. `4.7`
C. `9.9`
D. 8

Answer

B

Explanation

`m_(y)v_(y)=m_(alpha)v_(alpha)rArrV_(y)~~2.115xx10^(5)m//s`
`:.` TE released during an a decay of the nucleus X is.