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

To solve the problem, we need to identify the two hydrogen-like atoms \( A \) and \( B \) based on the given conditions. Let's break down the solution step by step.

### Step 1: Understanding the Energy Difference
The energy levels of hydrogen-like atoms can be expressed as:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.
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