`A` and `B` are two hydrogen like atoms such that `m_(B)=2m_(A)`. Also, the number of protons and neutrons in the two nuclei are equal. Given that difference of photon energy corresponding to the first Balmer lines emitted by `A` and `B` is `2.667eV`. Let `Z_(A)` and `Z_(B)` be the atomic numbers of `A` and `B` respectively.

Options

A. `Z_(A)=4`
B. `Z_(A)=1`
C. `Z_(B)=2`
D. `Z_(B)=8`

Answer

B, C

Explanation

`m_(B)=2m_(A)impliesZ_(B)=2Z_(A)`. Let `Z_(A)=Z, Z_(B)=2Z`
So, `E_(B)-E_(A)=(13.6xx5)/36xx3Z^(2)=5.667`
`implies Z=1`