When a piece of metal is illuminated by a monochromatic light of wavelength `lambda`, then stopping potential is `3V_s`. When the same surface is illuminated by the light of wavelength `2 lambda`, then stopping potential becomes `V_s`. The value of threshold wavelength for photoelectric emission will be

Options

A. `4 lambda`
B. `8 lambda`
C. `4/3 lambda`
D. `6 lambda`

Answer

To solve the problem, we will use the principles of the photoelectric effect and Einstein's equation. Let's break it down step by step.

### Step 1: Write down the equations for stopping potential
When a piece of metal is illuminated by monochromatic light, the stopping potential \( V_0 \) is related to the energy of the incident photons and the work function of the metal. The equation is given by:

\[ eV_0 = \frac{hc}{\lambda} - \phi_0 \]

Where:
- \( e \) is the charge of an electron,
- \( V_0 \) is the stopping potential,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light,
- \( \phi_0 \) is the work function of the metal.

### Step 2: Set up equations for the two cases
1. For the first case (\( \lambda \), stopping potential \( 3V_s \)):
\[ e(3V_s) = \frac{hc}{\lambda} - \phi_0 \quad \text{(Equation 1)} \]

2. For the second case (\( 2\lambda \), stopping potential \( V_s \)):
\[ eV_s = \frac{hc}{2\lambda} - \phi_0 \quad \text{(Equation 2)} \]

### Step 3: Rearranging the equations
From Equation 1:
\[ 3eV_s = \frac{hc}{\lambda} - \phi_0 \quad \Rightarrow \quad \phi_0 = \frac{hc}{\lambda} - 3eV_s \]

From Equation 2:
\[ eV_s = \frac{hc}{2\lambda} - \phi_0 \quad \Rightarrow \quad \phi_0 = \frac{hc}{2\lambda} - eV_s \]

### Step 4: Set the two expressions for \( \phi_0 \) equal
Setting the two expressions for \( \phi_0 \) equal gives:
\[ \frac{hc}{\lambda} - 3eV_s = \frac{hc}{2\lambda} - eV_s \]

### Step 5: Solve for \( V_s \)
Rearranging the equation:
\[ \frac{hc}{\lambda} - \frac{hc}{2\lambda} = 3eV_s - eV_s \]
\[ \frac{hc}{2\lambda} = 2eV_s \]
\[ V_s = \frac{hc}{4e\lambda} \quad \text{(Equation 3)} \]

### Step 6: Substitute \( V_s \) back into one of the equations
Substituting \( V_s \) from Equation 3 into Equation 2:
\[ \phi_0 = \frac{hc}{2\lambda} - e\left(\frac{hc}{4e\lambda}\right) \]
\[ \phi_0 = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} \]
\[ \phi_0 = \frac{2hc}{4\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda} \]

### Step 7: Relate \( \phi_0 \) to the threshold wavelength
The work function \( \phi_0 \) can also be expressed in terms of the threshold wavelength \( \lambda_0 \):
\[ \phi_0 = \frac{hc}{\lambda_0} \]

### Step 8: Set the two expressions for \( \phi_0 \) equal
\[ \frac{hc}{\lambda_0} = \frac{hc}{4\lambda} \]

### Step 9: Solve for \( \lambda_0 \)
Cancelling \( hc \) from both sides:
\[ \frac{1}{\lambda_0} = \frac{1}{4\lambda} \]
\[ \lambda_0 = 4\lambda \]

### Final Answer
The threshold wavelength for photoelectric emission is:
\[ \lambda_0 = 4\lambda \]

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