Question

Energy liberated in the de-excitation of hydrogen atom from $3^{rd}$ level to $1^{st}$ 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 $2^{nd}$ energy level, it is found that the de-Broglie wavelength of the fastest photoelectron, now ejected has decreased by a factor of 3. For this new gas, difference of energies of $2^{nd}$ Lyman line and $1^{st}$ Balmer line is found to be 3 times the ionization potential of the hydrogen atom. Select the correct statement(s):

• 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 1

Answer: (B), (C)

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

The energy emitted by the photoelectron is given by
$E=E_0{z}^2(\frac{1}{n_f^2}-\frac{1}{n_i^2})$
where $z=$ atomic number of gas
$E_0=$ ionization potential of hydrogen atom
So given that the differene between the energies is $3E_0$
$\therefore E_0{z}^2(1-\frac{1}{9})-E_0{z}^2(\frac{1}{4}=\frac{1}{9})=3E_0$
$z=2$
Let wavelength of the hydrogen atom be ${\lambda }_1$ and that of other be ${\lambda }_2$
So given that
${\lambda }_1/{\lambda }_2=3$
$K{E}_1=E_0(1-\frac{1}{9})-\varphi$ ...(i)
$K{E}_2=E_0{z}^2(1-\frac{1}{4})-\varphi$ ...(ii)
And since we know that
$KE\propto \frac{1}{\lambda }$
So now
As $\lambda$ becomes $(\lambda -\frac{\lambda }{3})=\frac{2\lambda }{3}$ as given in question
Then $KE=\frac{3KE}{2}$
Now putting this value in (i) and (ii) and solving we get
$\varphi =8.5\text{eV}$