Two hydrogen-like atoms `A` and `B` are of different masses and each atom contains equal numbers of protons and neutrons. The difference in the energies between the first Balmer lines emitted by `A and B` , is `5.667 e V`. When atom atoms `A and B` moving with the same velocity , strike a heavy target , they rebound with the same velocity in the process, atom `B` imparts twice the momentum to the target than that `A` imparts. Identify the atom `A and B`.

Options

A. `12.1 eV`
B. `13.6 eV`
C. `14.3 eV`
D. `15.1 eV`

Answer

D

Explanation

`Delta E = { 13.6 Z_(B)^(2) ((1)/(1^(2)) - (1)/(2^(2))) - 13.6 Z_(A)^(2) ((1)/(1^(2)) - (1)/(2^(2)))} eV`
`81.6 eV = (13.6 xx3)/(4) (Z_(B)^(2) - Z_(A)^(2)) = 8` (i)
Using conservation of momentum. For `A, m_(A) u = MV_(1) - m_(A) (u)/(2)`
`(3)/(2)m_(A) u = MV_(1)`
For `B, (3)/(2)m_(B) u = MV_(2)`
But `MV_(2) = 3 MV_(1)`
`m_(B) - 3m_(A)`
Since both `A and B` carry same number of proton and neutron , we have
`Z_(B) = 3Z_(A)`
But `Z_(B)^(2) - Z_(A)^(2) = 8`
`9 Z_(A)^(2), Z_(BN)^(2) = 8`
`Z_(A) = 1, Z_(B) = 3`
Hence ,`A` is `_(1)^(2) H` and `B` is `_(4)^(6)Li`
Now the difference in energy between the first Balmer lines emitted by `A and b`
`Delta E = 13.6 ((1)/(2^(2)) - (1)/(3^(2))) Z_(B)^(2) - 13.6 ((1)/(2^(2)) - (1)/(3^(2))) Z_(A)^(2)`
`= 13.6 xx (5)/(36) xx (Z_(B)^(2) - Z_(A)^(2))`
`= 13.6 xx 5 xx (8)/(36) = 15.1 eV`