A nucleus X, initially at rest, undergoes alpha-decay according to the equation.
`_92^AXrarr_Z^228Y+alpha`
(a) Find the values of A and Z in the above process.
(b) The alpha particle produced in the above process is found in move in a circular track of during the process and the binding energy of the parent nucleus X.
Given that `m(Y)=228.03u`, `m(`_0^1n)=1.009u`
`m(`_2^4He)=4.003u`, `m(`_1^1H)=1.008u`

Answer

A, B, C

Explanation

(a) `A-4=228`
:. `A=232`
`92-2=Z`
or `Z=90`
(b) From the relation,
`r=(sqrt(2Km))/(Bq) implies K_alpha = (r^2B^2q^2)/(2m)`
`=((0.11)^2(3)^2(2xx1.6xx10^-19)^2)/(2xx4.003xx1.67xx10^-27xx1.6xx10^-13)MeV`
`=5.21 MeV`
From the conservation of mementum,
`p_Y=p_alpha or sqrt(2K_Ym_Y)=sqrt(2K_alpham_alpha)`
:. `K_Y=((m_alpha)/(m_Y))K_alpha=4.003/228.03xx5.21`
`=0.09 MeV`
:. Total energy released `=K_alpha+K_Y=5.3MeV`
Total binding energy of daughter products
`=[92xx(mass of proton)+(232-92)(mass of neutron)-(m_Y)-(m_alpha)]xx931.48MeV`
`=[(92xx1.008)+(140)(1.009)-228.03-4.003]931.48MeV`
`=1828.5 MeV`
:. Binding energy of parent nucleus
`=binding energy of daughter products-energy released`
`=(1828.5-5.3)MeV=1823.2MeV`