When photones of energy 4.0 eV fall on the surface of a metal A, the ejected photoelectrons have maximum kinetic energy `T_(A)` ( in eV) and a de-Broglie wavelength `lambda_(A)`. When the same photons fall on the surface of another metal B, the maximum kinetic energy of ejected photoelectrons is `T_(B) = T_(A) -1.5eV`. If the de-Broglie wavelength of these photoelectrons is `lambda_(B) =2 lambda _(A)`, then the work function of metal B is

Options

A. 2eV
B. 3eV
C. 2.5 eV
D. 3.5 eV

Answer

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

### Step 1: Understand the given information
We know that:
- The energy of the incident photons, \( E = 4.0 \, \text{eV} \).
- The maximum kinetic energy of photoelectrons from metal A is \( T_A \).
- The maximum kinetic energy of photoelectrons from metal B is \( T_B = T_A - 1.5 \, \text{eV} \).
- The de-Broglie wavelength of photoelectrons from metal B is \( \lambda_B = 2 \lambda_A \).

### Step 2: Use the de-Broglie wavelength formula
The de-Broglie wavelength \( \lambda \) of a particle is given by:

\[ \lambda = \frac{h}{\sqrt{2mT}} \]

Where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the electron,
- \( T \) is the kinetic energy.

For metal A, we have:

\[ \lambda_A = \frac{h}{\sqrt{2mT_A}} \]

For metal B, we have:

\[ \lambda_B = \frac{h}{\sqrt{2mT_B}} \]

### Step 3: Relate the wavelengths
Given that \( \lambda_B = 2 \lambda_A \):

\[ \frac{h}{\sqrt{2mT_B}} = 2 \cdot \frac{h}{\sqrt{2mT_A}} \]

### Step 4: Simplify the equation
Cancel \( h \) and \( \sqrt{2m} \) from both sides:

\[ \frac{1}{\sqrt{T_B}} = \frac{2}{\sqrt{T_A}} \]

### Step 5: Square both sides
Squaring both sides gives:

\[ \frac{1}{T_B} = \frac{4}{T_A} \]

This implies:

\[ T_A = 4T_B \]

### Step 6: Substitute the expression for \( T_B \)
From the earlier relation \( T_B = T_A - 1.5 \, \text{eV} \), we substitute \( T_A \):

\[ T_B = 4T_B - 1.5 \]

### Step 7: Rearranging the equation
Rearranging gives:

\[ 1.5 = 4T_B - T_B \]

\[ 1.5 = 3T_B \]

### Step 8: Solve for \( T_B \)
Thus,

\[ T_B = \frac{1.5}{3} = 0.5 \, \text{eV} \]

### Step 9: Use Einstein's photoelectric equation
According to Einstein's photoelectric equation:

\[ T_B = E - \phi_B \]

Where \( \phi_B \) is the work function of metal B. Substituting the known values:

\[ 0.5 = 4 - \phi_B \]

### Step 10: Solve for \( \phi_B \)
Rearranging gives:

\[ \phi_B = 4 - 0.5 = 3.5 \, \text{eV} \]

### Final Answer
The work function of metal B is \( \phi_B = 3.5 \, \text{eV} \).

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