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 `

Explanation

The given equation is `._(92)^(A)X rarr ._(z)^(228) Y +._(2)^(4)He`
`A = 228 +4 = 232 , 92 = z+2 :. z = 90`
`(m_(alpha)v_(alpha)^(2))/(r ) = q v_(alpha)B, v_(alpha) = sqrt((e rB)/(m_(alpha)))`
`= sqrt((1.1 xx 10^(2) xx 2 xx 1.6 xx 10^(-19) xx 3xx 10^(3))/(4.003 xx 1.66 xx 10^(-27)))`
`= 4.0 xx 10^(6) m//s`
From conservation of linear momentum, `m_(alpha) v_(alpha) = m_(y)v_(y)`
`v_(y) = (m_(alpha)v_(alpha))/(m_(y)) = ((4.003)(4.0 xx 10^(6)))/((228.03)) = 7.0 xx 10^(4)m//s`
There fore, energy released during the process
`= (1)/(2)[m_(alpha)v_(alpha)^(2)+m_(y)v_(y)^(2)] = ((1.66 xx 10^(-27)))/((2 xx 1.6 xx 10^(-13)))`
`[(4.003)(4.0 xx 10^(6))^(2)+(228.03)(7.0 xx10^(4))^(2)]MeV`
`= 0.34 Me V = (0.34)/(931.5)am u = 0.000365 am u`
Therefore, mass of `._(92)^(232)X = m_(y)+m_(alpha) + 0.000365`
`= 232.033365 u`
Mass defect `Delta m = 92(1.008)+(232-92)(1.009)-232.033365`.
`:.` Binding energy `=1.9626635xx931.5 MeV`
`=1828.2MeV`