Question

Energy liberated in the de-excitation of hydrogen atom from the third level to the first 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 the second energy level, it is found that the de Broglie wavelength of the fastest photoelectrons now ejected has decreased by 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 answers.

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

Answer 1

Answer: (B), (C)

The correct options are
B The gas is helium
C The work function of photo-cathode is $8.5eV$

Second Lyman line: $n=3$ to $n=1$
First Balmer line: $n=3$ to $n=2$
Difference in their energies is: $E_2-E_1$
which is, $(13.6-3.4)Z^2$ and given that it is equal to 3 times I.E of hydrogen.
$\Rightarrow 10.2Z^2=3\times 13.6$.

Therefore, $Z=2$
Therefore, the gas is Helium.
Energy liberated by the de-excitation of hydrogen from $n=3$to$n=1$is,$13.6-1.5=12.1\ eV$

It is now exposed to helium excited to second energy level.
Energy released from helium is $E_2-E_1=40.8eV$
Second time, emitted photo electrons had de-Broglie wavelength decreased by factor $3$, which is $\lambda \to \frac{2\lambda }{3}\Rightarrow K.E\to \frac{3K.E}{2}$

Therefore,
$12.1=\Phi +K.E⟶\left(i\right)40.8=\Phi +\frac{3K.E}{2}⟶\left(ii\right)$
Solving these equations, we get $\Phi =8.5eV$.