Photon having energy equivalent to the binding energy of `4th` state of `He^(+)` atom is used to eject an electron from the metal surface of work function `1.4 eV`. If electrons are further accelerated through the potential difference of `4V` then the minimum value of De-brogile wavelength associated with the electron is:
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Correct Answer - 5
In the particular excited state of H-atom, path length is five times the de-Broglie wavelength.
`:. 2pir=5lambda …(1)`
However, path length in a sate `n` is `n` times the de-Broglie wavelength.
`:. 2pir=nlambda …(2)`
From (1) & (2), principal quantum number (n) of the excited state `=5`. Photon having `2^(nd)` highest energy corresponds to back transition of electron from `n=4` to `n=1`
This photon will cause an already excited `Li^(2+)` electron to go some higher state. Let the initial excited state of `Li^(2+)` ion be `n_(1)` and final excited state of `Li^(2+)` ion be `n_(2)`
`:. (13.6(1)^(2)[1/1^(2)-1/4^(2)])/(("Photon having" 2^(nd) "highest energy"),("corresponding to transition n=4 to n=1 in H-atom"))=(13.6(3)^(2)[1/n_(1)^(2)-1/n_(2)^(2)])/(("Energy absorbed by" Li^(+) "ion to make"),("a transition from" n_(1) "to" n_(2)))`
`:. 13.6[1/1^(2)-1/4^(2)]=13.6[3^(2)/n_(1)^(2)-3^(2)/n_(2)^(2)]` or `13.6[1/1^(2)-1/4^(2)]=13.6[1/((n_(1)//3)^(2))-1/((n_(2)//3)^(2))]`
On cmparing both sides,
`n_(1)/3=1` & `n_(2)/3=4rArr n_(1)=3` & `n_(2)=12`
Thus, the final excited state of `Li^(2+)` ion electron is `n=12` Ans.
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