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:

Photon having energy equivalent to the binding energy of 4th state of

1.	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:
Answer» Solution :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.