Photon having energy equivalent to the binding energy of `4^(th)` 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-broglie wavelength associated with the electron is :

Answer

5

Explanation

Total energy `=(13.6Z^2)/(n^2) =- (13.6 (Z)^2)/((4)^2) = 3.4 eV`
Now K.E. = 3.4 - 1.4 = 2eV
Now , total energy = 2+ 4 = 6 eV
i.e. potential = 6V
For electron `l= sqrt((150)/(V)) , ` So l = 5 Å