Photons of energy 7 eV are incident on two metals A and B with work functions 6 eV and 3 eV respectively. The minimum de Broglie wavelengths of the emitted photoelectrons with maximum energies are `lambda_A` and `lambda_B`, respectively where `lambda_A // lambda_B` is nearly

Options

A. 0.5
B. 14
C. `4.0`
D. `2.0`

Answer

To solve the problem, we need to find the ratio of the minimum de Broglie wavelengths of the emitted photoelectrons from metals A and B, denoted as \( \lambda_A \) and \( \lambda_B \), respectively. We will follow these steps:

### Step 1: Determine the Kinetic Energy of Photoelectrons
The energy of the incoming photons is given as \( E = 7 \, \text{eV} \). The work functions for metals A and B are \( \phi_A = 6 \, \text{eV} \) and \( \phi_B = 3 \, \text{eV} \), respectively.

The kinetic energy (KE) of the emitted photoelectrons can be calculated using the formula:
\[ \text{KE} = E - \phi \]

For metal A:
\[ \text{KE}_A = E - \phi_A = 7 \, \text{eV} - 6 \, \text{eV} = 1 \, \text{eV} \]

For metal B:
\[ \text{KE}_B = E - \phi_B = 7 \, \text{eV} - 3 \, \text{eV} = 4 \, \text{eV} \]

### Step 2: Relate Kinetic Energy to Velocity
The kinetic energy of an electron can also be expressed in terms of its mass and velocity:
\[ \text{KE} = \frac{1}{2} m v^2 \]

Thus, we can write:
\[ \text{KE}_A = \frac{1}{2} m v_A^2 \quad \text{and} \quad \text{KE}_B = \frac{1}{2} m v_B^2 \]

### Step 3: Set Up the Ratio of Kinetic Energies
Now, we can set up the ratio of the kinetic energies:
\[ \frac{\text{KE}_A}{\text{KE}_B} = \frac{1 \, \text{eV}}{4 \, \text{eV}} = \frac{1}{4} \]

### Step 4: Relate the Velocities
Using the relationship between kinetic energy and velocity, we have:
\[ \frac{\frac{1}{2} m v_A^2}{\frac{1}{2} m v_B^2} = \frac{1}{4} \]
Since \( \frac{1}{2} m \) cancels out, we get:
\[ \frac{v_A^2}{v_B^2} = \frac{1}{4} \]
Taking the square root gives:
\[ \frac{v_A}{v_B} = \frac{1}{2} \]

### Step 5: Calculate the de Broglie Wavelengths
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
Thus, we can write:
\[ \lambda_A = \frac{h}{m v_A} \quad \text{and} \quad \lambda_B = \frac{h}{m v_B} \]

### Step 6: Set Up the Ratio of Wavelengths
Now, we can find the ratio of the wavelengths:
\[ \frac{\lambda_A}{\lambda_B} = \frac{\frac{h}{m v_A}}{\frac{h}{m v_B}} = \frac{v_B}{v_A} \]

Substituting the velocity ratio we found earlier:
\[ \frac{\lambda_A}{\lambda_B} = \frac{v_B}{v_A} = 2 \]

### Final Result
Thus, the ratio of the minimum de Broglie wavelengths of the emitted photoelectrons is:
\[ \frac{\lambda_A}{\lambda_B} \approx 2 \]