Energy liberated in the de-excitation of hydrogen atom from the third level falls on a photo-cathode. Later when the same photo-cathode is exposed 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`.

Explanation

To solve the problem step by step, we will analyze the information provided and apply relevant physics concepts.

### Step 1: Understanding the Energy Levels
We know that the energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For the hydrogen atom, the energy levels for \( n = 1, 2, 3 \) are:
- \( E_1 = -13.6 \, \text{eV} \)
- \( E_2 = -3.4 \, \text{eV} \)
- \( E_3 = -1.51 \, \text{eV} \)

### Step 2: Energy Difference for the Second Lyman Line and First Balmer Line
The second Lyman line corresponds to the transition from \( n = 3 \) to \( n = 1 \):
\[ E_{L2} = E_3 - E_1 = (-1.51) - (-13.6) = 12.09 \, \text{eV} \]

The first Balmer line corresponds to the transition from \( n = 3 \) to \( n = 2 \):
\[ E_{B1} = E_3 - E_2 = (-1.51) - (-3.4) = 1.89 \, \text{eV} \]

### Step 3: Energy Difference
The difference in energies between the second Lyman line and the first Balmer line is:
\[ E_{L2} - E_{B1} = 12.09 \, \text{eV} - 1.89 \, \text{eV} = 10.2 \, \text{eV} \]

### Step 4: Relating to Ionization Potential
We are given that this energy difference is three times the ionization potential of hydrogen:
\[ E_{L2} - E_{B1} = 3 \times 13.6 \, \text{eV} \]
Thus:
\[ 10.2 \, \text{eV} = 3 \times 13.6 \, \text{eV} \]
This leads us to find the value of \( Z \) for the unknown gas.

### Step 5: Finding \( Z \)
Using the energy difference formula for the unknown gas:
\[ E_{L2} - E_{B1} = 10.2 Z^2 \, \text{eV} \]
Setting this equal to \( 3 \times 13.6 \):
\[ 10.2 Z^2 = 40.8 \]
Solving for \( Z^2 \):
\[ Z^2 = \frac{40.8}{10.2} = 4 \implies Z = 2 \]
This indicates the unknown gas is helium (He).

### Step 6: De Broglie Wavelength and Kinetic Energy
The de Broglie wavelength \( \lambda \) is inversely proportional to the kinetic energy \( KE \):
\[ \lambda \propto \frac{1}{KE} \]
If the wavelength decreases by a factor of 3, the kinetic energy increases by a factor of 3:
\[ KE' = 3 \times KE \]

### Step 7: Applying Einstein's Photoelectric Equation
Using the photoelectric equation:
\[ KE = E_{photon} - \phi \]
Where \( \phi \) is the work function. For hydrogen:
\[ KE = 12.1 - \phi \]
For helium:
\[ KE' = 40.8 - \phi \]

### Step 8: Setting Up the Equations
From the previous steps, we have:
1. \( KE = 12.1 - \phi \)
2. \( 3 \times KE = 40.8 - \phi \)

Substituting \( KE \) from the first equation into the second:
\[ 3(12.1 - \phi) = 40.8 - \phi \]
Expanding and simplifying:
\[ 36.3 - 3\phi = 40.8 - \phi \]
\[ 2\phi = 40.8 - 36.3 \]
\[ 2\phi = 4.5 \implies \phi = 2.25 \, \text{eV} \]

### Conclusion
The work function \( \phi \) of the material is \( 2.25 \, \text{eV} \).