Silver has a work function of 4.7 eV. When ultraviolet light of wavelength 100 mm is incident upon it, a potential of 7.7 V is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop the photoelectrons when light of wavelength 200mm is incident upon silver?

Options

A. 1.5 V
B. 3.85 V
C. 2.35 V
D. 15.4 V

Explanation

To solve the problem step by step, we will use the concepts from the photoelectric effect and the given data.

### Step 1: Understand the given data
- Work function of silver (W) = 4.7 eV
- Wavelength of the first light (λ1) = 100 nm
- Stopping potential for the first light (V0) = 7.7 V
- Wavelength of the second light (λ2) = 200 nm

### Step 2: Calculate the energy of the photon for the first wavelength
The energy of a photon (E) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 4.1357 \times 10^{-15} \) eV·s
- \( c \) (speed of light) = \( 3 \times 10^8 \) m/s
- \( \lambda \) (wavelength in meters) = \( 100 \, \text{nm} = 100 \times 10^{-9} \, \text{m} \)

Substituting the values:
\[ E_1 = \frac{(4.1357 \times 10^{-15} \, \text{eV·s})(3 \times 10^8 \, \text{m/s})}{100 \times 10^{-9} \, \text{m}} \]
\[ E_1 = \frac{1.241 \times 10^{-6} \, \text{eV·m}}{100 \times 10^{-9} \, \text{m}} \]
\[ E_1 = 12.41 \, \text{eV} \]

### Step 3: Relate the stopping potential to the energy of the photon
According to the photoelectric effect:
\[ eV_0 = E - W \]
Where \( e \) is the charge of an electron (which cancels out in the equation). For the first case:
\[ 7.7 \, \text{eV} = E_1 - 4.7 \, \text{eV} \]
\[ E_1 = 7.7 \, \text{eV} + 4.7 \, \text{eV} = 12.4 \, \text{eV} \]
This confirms our calculation for \( E_1 \).

### Step 4: Calculate the energy of the photon for the second wavelength
Using the relationship between the wavelengths and energies:
\[ \frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} \]
Substituting the values:
\[ E_2 = E_1 \cdot \frac{\lambda_1}{\lambda_2} \]
\[ E_2 = 12.4 \, \text{eV} \cdot \frac{100 \, \text{nm}}{200 \, \text{nm}} \]
\[ E_2 = 12.4 \, \text{eV} \cdot 0.5 = 6.2 \, \text{eV} \]

### Step 5: Calculate the stopping potential for the second wavelength
Using the same relationship:
\[ eV_{02} = E_2 - W \]
\[ V_{02} = \frac{E_2 - W}{e} \]
Substituting the values:
\[ V_{02} = 6.2 \, \text{eV} - 4.7 \, \text{eV} \]
\[ V_{02} = 1.5 \, \text{V} \]

### Final Answer
The potential required to stop the photoelectrons when light of wavelength 200 nm is incident upon silver is **1.5 V**.

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