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

(i) Given decay equation can be written as follows:
`""_(92)X^(A) to ""_(2)He^(4)`.
`A = 228 + 4 = 232`
`92 = Z +2 rArr Z =90`
Let alpha-particle be emitted with speed `v_(alpha)` perpendicular to magnetic field B, then magnetic force acting on it acts as centripetal force for circular motion, hence we can write the following.
`(2e)v_(alpha)B =(m_(alpha)v_(alpha)^(2))/r`
In the above equation, ma is mass of alpha particle and r is radius of its circular path. From the above equation, we can write speed of the alpha particle as follows:
`v_(alpha) = (2erB)/m_(alpha)`
`rArr v_(alpha) =(2 xx 1.6 xx 10^(-19) xx 0.11 xx 3)/(4.003 xx 1.66 xx 10^(-27))`
`rArr v_(alpha) =1.59 xx 10^(7)` m/s
If `m_(y)` is mass of daughter nucleus Y and velocity with which it recoils is `v_(y)`, then according to conservation of linear momentum, we can write the following.
`m_(y)v_(y)=m_(alpha)v_(alpha)`
`rArr v_(Y) =(m_(alpha) v_(alpha))/(m_(y))`
n the given decay process, energy released is equal to kinetic energy possessed by alpha particle and daughter nucleus Y. Hence total energy released can be written as follows:
`E = 1/2m_(alpha)v_(alpha)^(2) + 1/2m_(Y)v_(y)^(2)`
`rArr E = 8.547 xx 10^(-13)` joules.
`rArr E=(8.547 xx 10^(-12))/(1.6 xx 10^(-13)) MeV = 5.34 MeV`
(ii) Mass defect of the given process to produce above calculated amount of energy can be written as follows:
`Deltam = 5.34/931 =0.00574` amu
Mass defect according to decay equation can be written as follows:
`Deltam =m_(X)-m_(Y)-m_(alpha)`
`rArr 0.00575 = m_(X)-228.03-4.003`
`rArr m_(X) = 232.03875` amu
The above-calculated mass of nucleus X is actual. On the other hand, we can calculate expected mass of nucleus X according to its number of nucleons. Difference between expected mass and actual mass is nuclear mass defect for X. By multiplying nuclear mass defect with 931 `MeV//c^(2))` , we can find binding energy for nucleus X.
Nucleus X has 92 protons and 232-92 = 140 neutrons.
Hence expected mass of the nucleus.
`X = 92 xx 1.008 + 140 xx 1.009 = 233.996` amu
Hence nuclear mass defect
`Deltam = 233.996 - 232.03875 = 1.957` amu
Equivalent amount of energy which is binding energy of nucleus can be written as follows:
`DeltaE = 1.957 xx 931 = 1822` MeV.
Hence binding energy of nucleus X is 1822 MeV.