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 equatin is
`"" _(92) ^(A) X to "" _(z) ^(228) Y + "" _(2) ^(4) He`
`A = 228 + 4 = 232 , 92 = z + 2 therefore z = 90`
`(m _(a) v _(a) ^(2))/( r ) = q v _(a) B, v _(a) = sqrt((r q B)/(m_(a)))`
`= sqrt((1.1 xx 10 ^(2) xx 2 xx 1.6 xx 10 ^(-19) xx 3 xx 10 ^(3))/(4. 003 xx 1. 66 xx 10 ^(-27)))`
`= 4.0 xx 10 ^(6) m //s `
From conservation of linear momentum,
`m _(a) v _(a) = 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)]`
` = 1/2 [(4.003) (4.0 xx 10 ^(6)) ^(2) +`
`(228.03) (7.0 xx 10 ^(6)) ^(2) ]xx`
`(1.66 xx 10 ^(-27))/( 1. 6 xx 10 ^(-13)) MeV`
`=0.34 MeV = (0.34)/(931.5) am u = 0.000 365 am u`
Therefore, mass of
`"" _(92) ^(232) X = m _(y ) + m _(a) + 0.000365`
` = 232 . 0333 65u`
Mass defect
`Delta m = 92 ( 1. 008) + (232- 92) `
` ( 1. 009) - 232 . 033365= 1. 962635u`
`therefore` Binding energy
`= 1. 96 2635 x 931.5 Mev`
`=1828.2 Mev`
` = 1828.2 Mev`