When a metal surface is illuminated by light of wavelength `lambda` ,the stopping potential is "8V" .When the same surface is illuminated by light of wavelength `3 lambda` , stopping potential is "2V" .The threshold wavelength for this surface is:

Options

A. `3lamda`
B. `5lamda`
C. `9lamda`
D. `4.5lamda`

Answer

To find the threshold wavelength for the metal surface illuminated by light of wavelength \( \lambda \) and \( 3\lambda \), we can use the photoelectric effect equation derived from Einstein's photoelectric equation:

\[ E = h \nu = \frac{hc}{\lambda} \]

Where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the light.

The stopping potential \( V_0 \) is related to the maximum kinetic energy of the emitted electrons:

\[ eV_0 = E - E_0 \]

Where \( E_0 \) is the work function of the metal. The stopping potential can be expressed in terms of the wavelength of the incident light.

### Step 1: Write the equations for the two conditions

1. For the wavelength \( \lambda \) with stopping potential \( V_0 = 8V \):
\[ e \cdot 8 = \frac{hc}{\lambda} - \phi \quad \text{(Equation 1)} \]

2. For the wavelength \( 3\lambda \) with stopping potential \( V_0 = 2V \):
\[ e \cdot 2 = \frac{hc}{3\lambda} - \phi \quad \text{(Equation 2)} \]

### Step 2: Rearrange both equations

From Equation 1:
\[ \phi = \frac{hc}{\lambda} - 8e \]

From Equation 2:
\[ \phi = \frac{hc}{3\lambda} - 2e \]

### Step 3: Set the two expressions for \( \phi \) equal to each other

\[ \frac{hc}{\lambda} - 8e = \frac{hc}{3\lambda} - 2e \]

### Step 4: Solve for \( hc/\lambda \)

Rearranging gives:
\[ \frac{hc}{\lambda} - \frac{hc}{3\lambda} = 8e - 2e \]

This simplifies to:
\[ \frac{hc}{\lambda} - \frac{hc}{3\lambda} = 6e \]

Combining the left side:
\[ \frac{3hc - hc}{3\lambda} = 6e \]
\[ \frac{2hc}{3\lambda} = 6e \]

Multiplying both sides by \( 3\lambda \):
\[ 2hc = 18e\lambda \]
\[ hc = 9e\lambda \quad \text{(Equation 3)} \]

### Step 5: Substitute \( hc \) back into one of the equations for \( \phi \)

Using Equation 1:
\[ \phi = \frac{9e\lambda}{\lambda} - 8e \]
\[ \phi = 9e - 8e = e \]

### Step 6: Find the threshold wavelength

The threshold wavelength \( \lambda_0 \) is given by:
\[ \phi = \frac{hc}{\lambda_0} \]

Substituting \( \phi = e \):
\[ e = \frac{hc}{\lambda_0} \]

Using \( hc = 9e\lambda \):
\[ e = \frac{9e\lambda}{\lambda_0} \]

Cancelling \( e \) from both sides:
\[ 1 = \frac{9\lambda}{\lambda_0} \]

Thus:
\[ \lambda_0 = 9\lambda \]

### Final Answer
The threshold wavelength for the surface is \( 9\lambda \).

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