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Q.

Energy liberated in the de–excitation of hydrogen atom from 3rd level to 1st level falls on a photo–cathode. Later when the same photo–cathode is exposed to a spectrum of some unknown hydrogen like gas, excited to 2nd 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  2nd Lyman line and 1st Balmer line found to be 3 times the ionization energy of the hydrogen atom. Select the correct statement(s):
 

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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 is B, C.

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Detailed Solution

E0z2(119)E0z2(1419)=3E0 z = 2 λ1λ2=3 KE1=E0(119)ϕ KE2=E0z2(114)ϕ KE1λ2=8.5eV

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