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

Let `Z_(A) and Z_(B)` be the atomic numbers and `m_(A) and m_(B)` be the mass numbers of hydrogen-like atoms `A and B`, respectively . Energy of nth state of hydrogen-like atoms is
`E_(n) = - (Z^(2) Rhc)/(n^(2)) = - (Z^(2) xx 13.6)/(n^(2)) e V`
Energy emitted for 1 line of Balmer series for atom`A`.
` Delta E_(1) = - Z_(A)^(2) xx 13.6 ((1)/(2^(2)) - (1)/(3^(2))) e V`
Energy emitted for 1 line of Balmer series for atom`B`.
` Delta E_(2) = Z_(B)^(2) xx 13.6 ((1)/(2^(2)) - (1)/(3^(2))) e V`
Given ` Delta E_(1) - Delta E_(2) = 5.667 e V, therefore
`5.667 e V = (Z_(B)^(2) - Z_(A)^(2)) xx 13.6 ((1)/(2^(2)) - (1)/(3^(2))) e V`
`(Z_(B)^(2) - Z_(A)^(2)) = (5.667 xx 36)/(13.6 xx 5) = 3` (i)
Let `u` be the intial velocities of each from `A` and `B` , respectively, then according to principal convervation of momentum for `A`
`m_(A) u = M v_(1) - m_(A) u`
or `M v_(1) - 2 m_(A) u` (ii)
Similarly, for `B M v_(2) = 2 m_(B) u` (iii)
Given , `M v_(2) = 2 mv_(1)`, therefore.
`2 m_(B) u = 2 (2 m_(A) u)`
This gives , ` `m_(B) = 2 m_(A)` (iv)
As number of protons and neutrons in each of `A and B` same separately .
`:. m_(B) = 2 Z_(B)` and `m_(A) = 2 Z_(A)` (v)
Substituting this in (iv) , we get
`2 Z_(B) = 2 (2 Z_(A)), i.e., Z_(B) = 2 Z_(A)`,
Solving Eq. (i) and Eq. (v), we get `Z_(A) = 1` and `Z_(B) = 2` i.e., atom `B` is singly ionized helium.