Two photons of energy 4eV and 4.5 eV incident on two metals A and B respectively. Maximum kinetic energy for ejected electron is `T_A` for A and `T_B = T_A - 1.5 eV` for metal B. Relation between de-Broglie wavelength of ejected electron of A and B are `lambda_B = 2lambda_A` . The work function of metal B is–

Options

A. 3eV
B. 1.5eV
C. 4eV
D. 2eV

Answer

To find the work function of metal B, we will follow these steps:

### Step 1: Define the given data
We have two photons with energies:
- Photon incident on metal A: \( E_A = 4 \, \text{eV} \)
- Photon incident on metal B: \( E_B = 4.5 \, \text{eV} \)

The maximum kinetic energies of the ejected electrons are:
- For metal A: \( T_A \)
- For metal B: \( T_B = T_A - 1.5 \, \text{eV} \)

### Step 2: Relate the de-Broglie wavelengths
We know that the relationship between the de-Broglie wavelengths of the ejected electrons is given by:
\[ \lambda_B = 2 \lambda_A \]

### Step 3: Use the de-Broglie wavelength formula
The de-Broglie wavelength \( \lambda \) of a particle is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the particle. The momentum can also be expressed in terms of kinetic energy \( T \):
\[ p = \sqrt{2mT} \]
Thus, we can write:
\[ \lambda = \frac{h}{\sqrt{2mT}} \]

### Step 4: Set up the ratio of wavelengths
From the relationship \( \lambda_B = 2 \lambda_A \), we can express this in terms of kinetic energies:
\[ \frac{\lambda_A}{\lambda_B} = \frac{1}{2} \implies \frac{\sqrt{2mT_B}}{\sqrt{2mT_A}} = \frac{1}{2} \]
Squaring both sides gives:
\[ \frac{T_B}{T_A} = \frac{1}{4} \]

### Step 5: Substitute \( T_B \)
Substituting \( T_B = T_A - 1.5 \) into the equation:
\[ \frac{T_A - 1.5}{T_A} = \frac{1}{4} \]
Cross-multiplying yields:
\[ 4(T_A - 1.5) = T_A \]
Expanding this gives:
\[ 4T_A - 6 = T_A \]
Rearranging leads to:
\[ 3T_A = 6 \implies T_A = 2 \, \text{eV} \]

### Step 6: Calculate \( T_B \)
Now substituting back to find \( T_B \):
\[ T_B = T_A - 1.5 = 2 - 1.5 = 0.5 \, \text{eV} \]

### Step 7: Find the work function of metal B
The work function \( \phi_B \) can be found using the equation:
\[ T_B = E_B - \phi_B \]
Substituting the known values:
\[ 0.5 = 4.5 - \phi_B \]
Rearranging gives:
\[ \phi_B = 4.5 - 0.5 = 4 \, \text{eV} \]

### Final Answer
The work function of metal B is \( \phi_B = 4 \, \text{eV} \).

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