Energy liberated in the de-excitation of hydrogen atom from the third level falls on a photo-cathode. Later when the same hoto-cathode is expoced to a spectrum of some unknown - like gas, excited to the second energy level , it is found that the de Broglie wavelength of the fastest photoelectrons now ejected has decreased by a factor of `3`. For this new gas, difference of energies of the second Lyman line and the first balmer line is found to be `3` times the ionization potential of the hydrogen atom. Select the correct statement (s):

Options

A. The gas is lithium.
B. The gas is helium.
C. The work function of photo-cathode is `8.5 eV`.
D. The work function of photo-cathode is `5.5 eV`.

Answer

B, C

Explanation

`E_(0) z^(2) (1 - (1)/(9)) - E_(0) z^(2) ((1)/(4) - (1)/(9)) = 3 E_(0) implies (27)/(36) E_(0) z^(2) = 3 E_(0)`
`implies z = 2 (lambda_(1))/(lambda_(2)) : = 3`
`KE_(1) = E_(0)(1 - (1)/(9)) = phi`
`KE_(2) = E_(0)z^(2) (1 - (1)/(4)) = phi`
`KE prop (1)/(lambda_(2)) = 8.5 eV`