Photon having wavelength `lambda = 3.5xx10^(-7)m` and `lambda = 5.4xx10^(-7)m` incident on a metal surface successively, let in both cases the ration of their stopping potential is 2 : 1, find work function :

Options

A. 1.05 eV
B. 1.25 eV
C. 1.75 eV
D. 2.12 eV

Explanation

To solve the problem, we need to find the work function of the metal surface given the stopping potentials for two different wavelengths of incident photons. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
We have two wavelengths of photons:
- \( \lambda_1 = 3.5 \times 10^{-7} \, \text{m} \)
- \( \lambda_2 = 5.4 \times 10^{-7} \, \text{m} \)

The ratio of their stopping potentials is given as:
\[ \frac{V_0^1}{V_0^2} = 2 : 1 \]
This means:
\[ V_0^1 = 2V_0^2 \]

### Step 2: Use the Photoelectric Equation
The stopping potential \( V_0 \) is related to the energy of the incident photons and the work function \( \phi \) of the metal by the equation:
\[ eV_0 = E - \phi \]
where \( E \) is the energy of the photon given by:
\[ E = \frac{hc}{\lambda} \]
Here, \( h \) is Planck's constant and \( c \) is the speed of light.

### Step 3: Write the Equations for Each Wavelength
For \( \lambda_1 \):
\[ eV_0^1 = \frac{hc}{\lambda_1} - \phi \]
For \( \lambda_2 \):
\[ eV_0^2 = \frac{hc}{\lambda_2} - \phi \]

### Step 4: Substitute the Stopping Potential Ratio
From the ratio \( V_0^1 = 2V_0^2 \), we can express \( V_0^1 \) in terms of \( V_0^2 \):
\[ e(2V_0^2) = \frac{hc}{\lambda_1} - \phi \]
Substituting \( V_0^2 \) from the second equation:
\[ e(2V_0^2) = \frac{hc}{\lambda_1} - \phi \]
\[ eV_0^2 = \frac{hc}{\lambda_2} - \phi \]

### Step 5: Rearranging the Equations
Now we can rearrange both equations:
1. From \( eV_0^1 \):
\[ e(2V_0^2) + \phi = \frac{hc}{\lambda_1} \]
2. From \( eV_0^2 \):
\[ eV_0^2 + \phi = \frac{hc}{\lambda_2} \]

### Step 6: Eliminate \( \phi \)
Now, let's eliminate \( \phi \) by subtracting the second equation from the first:
\[ e(2V_0^2) + \phi - (eV_0^2 + \phi) = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} \]
This simplifies to:
\[ eV_0^2 = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} \]
\[ eV_0^2 = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]

### Step 7: Solve for \( \phi \)
Now substituting \( eV_0^2 \) back into one of the equations to find \( \phi \):
\[ \phi = \frac{hc}{\lambda_2} - eV_0^2 \]
Substituting \( eV_0^2 \) from above:
\[ \phi = \frac{hc}{\lambda_2} - hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]
\[ \phi = hc \left( \frac{1}{\lambda_2} - \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \right) \]
\[ \phi = hc \left( \frac{1}{\lambda_2} + \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right) \]
\[ \phi = hc \left( \frac{2}{\lambda_2} - \frac{1}{\lambda_1} \right) \]

### Step 8: Calculate \( \phi \)
Now we can substitute the values of \( h \) and \( c \):
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)

Substituting \( \lambda_1 \) and \( \lambda_2 \):
\[ \phi = 6.626 \times 10^{-34} \times 3 \times 10^8 \left( \frac{2}{5.4 \times 10^{-7}} - \frac{1}{3.5 \times 10^{-7}} \right) \]

### Step 9: Final Calculation
Calculating the above expression will yield the work function \( \phi \).