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`.

Explanation

`Z_(A)`: number of protons in A
`Z_(B)` : number of protons in B
`m_(A)=2 Z_(A), m_(B)=2Z_(B)`
`|13.6 Z_(A)^(B)((1)/(2^(2))-(1)/(3^(2)))-13.6 Z_(B)^(2)((1)/(2^(2))-(1)/(3^(2)))|=5.667`
`|Z_(A)^(2)-Z_(B)^(2)|=3`
Momentum imparted to target by `A=2 m_(A)u`
Momentum imparted to target by `B=2 m_(B)u`
`2 m_(g)u=2xx2 m_(A)u`
`m_(B)=2 m_(A)`
`Z_(B)=2Z_(A)`
`Z_(B)^(2)-Z_(A)^(2)=3`
`4Z_(A)^(2)-Z_(A)^(2)=3 implies Z_(A)=1. Z_(B)=2`
`A : ._(1)H^(2)`
`B : ._(2)H_(e)^(4)`