A monochromatic beam of light having photon energy `12.5 eV` is incident on a simple `A` of atomic hydrogen gas in which all almost are in the ground state. The emission spectra obtained from this sample is incident on another sample `B` of atomic hydrogen gas in which all atoms are in the first excited state.
Based on above information, answer the following question:
The atoms of sample `B`

Options

A. will ionize when emission spectra of `A` is incident on `B`
B. may ionize when emission spectra of `A` is incident on `B`
C. will be excited to some higher state but won't ionize
D. None of above.

Answer

D

Explanation

As photon energy of incident light is not equal to energy difference between two energy state, that is why sample `A` does not absorbe any photon and therefore remain in ground state, and hence , no emission spectra. Thus sample `B` remain as it is, but it will be de-excite itself to ground state by emitting radiations of energy `10.2 eV`

when photon energy is replaced by electron beam, then atom of sample `A` can absorb either `10.2 eV` ("to reach" `1^(st)` excited state) of 12.1 eV (to reach `2^(nd)` excited state) In emission spectra `A` , we will have `3` time corresponding to `n = 2 to 1, n = 3 to 2 and n = 3 to1` having energies equal to `10.2 eV, 1.9 eV, 12.1 eV`, respectively
As least energy of this emission spectra is corresponding to transition from `n = 2` to `3` and sample B is in `1^(st)` excited state `B` can be excited to some higher energy level and if `B` absorbs energy corresponding to `n = 1` to `2` and `n = 1 to 3`, then it may ionize.