The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is :

Options

A. 329 nm
B. 309 nm
C. 382 nm
D. 400 nm

Answer

To solve the problem step by step, we will use the photoelectric effect equations. The stopping potential is related to the energy of the incident photons and the work function of the material.

### Step 1: Identify the known values for the first case
- Wavelength (\( \lambda_1 \)) = 491 nm
- Stopping potential (\( V_0 \)) = 0.710 V

### Step 2: Convert the wavelength from nanometers to meters
\[ \lambda_1 = 491 \, \text{nm} = 491 \times 10^{-9} \, \text{m} \]

### Step 3: Use the equation for stopping potential
The energy of the incident photons can be expressed as:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) = Planck's constant \( = 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) = Speed of light \( = 3 \times 10^8 \, \text{m/s} \)

### Step 4: Calculate the energy of the incident photons for the first case
Using \( h \) and \( c \):
\[ E_1 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{491 \times 10^{-9}} \]
Calculating this gives:
\[ E_1 \approx 4.05 \times 10^{-19} \, \text{J} \]

### Step 5: Convert the energy to electron volts
To convert joules to electron volts, use the conversion \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E_1 \approx \frac{4.05 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.53 \, \text{eV} \]

### Step 6: Relate the stopping potential to the work function
The stopping potential is related to the work function (\( \phi \)) and the energy of the photons:
\[ E_1 = \phi + eV_0 \]
Where \( e \) is the charge of an electron. Rearranging gives:
\[ \phi = E_1 - eV_0 \]
Substituting the values:
\[ \phi = 2.53 \, \text{eV} - 0.710 \, \text{eV} = 1.82 \, \text{eV} \]

### Step 7: Identify the known values for the second case
- Stopping potential (\( V_0' \)) = 1.43 V
- Work function (\( \phi \)) remains the same = 1.82 eV

### Step 8: Use the equation for stopping potential for the second case
For the new wavelength (\( \lambda_2 \)):
\[ E_2 = \phi + eV_0' \]
Substituting the known values:
\[ E_2 = 1.82 \, \text{eV} + 1.43 \, \text{eV} = 3.25 \, \text{eV} \]

### Step 9: Relate the energy to the new wavelength
Using the equation:
\[ E_2 = \frac{hc}{\lambda_2} \]
Rearranging gives:
\[ \lambda_2 = \frac{hc}{E_2} \]

### Step 10: Calculate the new wavelength
Substituting the values:
\[ \lambda_2 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{3.25 \times 1.6 \times 10^{-19}} \]
Calculating this gives:
\[ \lambda_2 \approx 382.7 \, \text{nm} \]

### Final Answer
The new wavelength is approximately **382.7 nm**.

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