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 `A` after passing of light through it

Options

A. may be in the first excited state
B. may be in the second excited state
C. may be in both first and second excited state
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` to`1` 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.