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Two hydrogen-like atoms A and B are of d...

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`.

A

`""_(3)^(2)H`

B

`""_(2)^(1)H`

C

`""_(2)^(3)H`

D

`""_(1)^(2)H`

Text Solution

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The correct Answer is:
B
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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.