Two hydrogen like atoms `A` and `B` having `(N)/(Z)` ratio equal to 1 have atomic number `Z_(A)` and `Z_(B)` A is moving, while `B` is stationary and both are in ground state. The minimum kinetic energy which `A` must have to have an inelastic collision with B such that both `A` and `B` getting excited is `221 e V`. If situation is reversed and B is made to stike `A` which is stationary and both in ground state then minimum kinetic energy which B must have to make an inelastic coliision with both getting excited is `331.5 e V`. Assume mase of proton = mass of neutron.
The atomic formula of`A` and `B` are :-

Options

A. `_(1).^(1)H & _(2).^(4)He`
B. `_(2).^(4)He & _(3).^(7)Li`
C. `_(4).^(8)Be & _(6).^(12)C`
D. `_(2).^(4)"He" & _(3).^(6)Li`

Answer

D

Explanation

`m_(A) = 2Z_(A)m`
`m_(B) = 2Z_(B)m`
`(1)/(2) (2Z_(A)Z_(B))/(Z_(A)+Z_(B))mv^(2) = E`
`(1)/(2)(2Z_(A)m)v^(2) = (Z_(A)+Z_(B))/(Z_(B))E`
So `(221)/(Z_(B)) = (331.5)/(Z_(B))`
`(Z_(B))/(Z_(A)) = 1.5 = (3)/(2)`
Also
`(221) (Z_(B))/(Z_(A)+Z_(B)) = (13.6)(Z_(A)^(2)+Z_(B)^(2)) (3)/(4)`
`(16.25xx4)/(3) = ((Z_(A)^(2)+Z_(B)^(2))(Z_(A)+Z_(B)))/(Z_(B))`
`(65)/(3) = (Z_(A)^(2)+(9)/(4)Z_(A)^(2)) (1+(2)/(3))`
`(65)/(3) = (13Z_(A)^(2))/(4)xx(5)/(3)`
` Z_(A)^(2) = 4 impliesZ_(A) = 2&Z_(B)=3`