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 `

Answer

(a) `232,90`
(b) `5.3 MeV, 1823 MeV`

Explanation

(a) `A-4=228`
`:. A=232`
`92-2=Z`
(b) From the relation,
`r=(sqrt(2Km))/(Bq)`
`K_(alpha)=(r^(2)B^(2)q^(2))/(2m)=((0.11)^(2)(3)^(2)(2xx1.6xx10^(-19))^(2))/(2xx4.003xx1.67xx10^(-27)xx1.6xx10^(-13))MeV=5.21 MeV`
From the conservation of momentum,
`P_(gamma)=P_(alpha)`
or `sqrt(2k_(gamma)m_(gamma))=sqrt(2k_(alpha)m_(alpha)) " " :. K_(g)=((m_(alpha))/(m_(gamma)))k_(alpha)=(4.003)/(228.03)xx5.21=0.09 MeV`
`:.` Total energy released `=K_(alpha)+K_(gamma)=5.3 MeV`
Total binding energy of daughter products
`=[(92xx("mass of proton")+(232-92)("mass of neutron")-(m_(gamma))-m_(alpha)]xx931.4 MeV`
`=[(92xx1.008)+(140)(1.009)-228.03-4.003]931.48 MeV`
`=1828.5 MeV`
`:.` Binding energy of parent nucleus
`=`binding energy of daughter products`-`energy released
`=(1828.5-5.3)MeV=1823.2MeV`