Using light of wavelength 6000 Å stopping potential is obtained 2.4 volt for photo electric cell. If light of wavelength 4000 Å is used then stopping potential would be -

Options

A. 2.9 V
B. 1.9 V
C. 3.43 V
D. 9.4 V

Answer

To solve the problem, we will use the photoelectric effect equation and the relationship between stopping potential, incident energy, and work function.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect Equation**:
The maximum kinetic energy (K.E.) of the emitted electrons is given by:
\[ K.E. = E - \phi \]
where \( E \) is the energy of the incident photons and \( \phi \) is the work function of the metal.

2. **Calculating Energy for Wavelength 6000 Å**:
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 4.14 \times 10^{-15} \) eV·s), \( c \) is the speed of light (\( 3 \times 10^8 \) m/s), and \( \lambda \) is the wavelength in meters. However, we can use a simplified formula:
\[ E = \frac{12400 \text{ eV·Å}}{\lambda \text{ (in Å)}} \]
For \( \lambda = 6000 \) Å:
\[ E_1 = \frac{12400}{6000} = 2.0667 \text{ eV} \]

3. **Using Stopping Potential to Find Work Function**:
The stopping potential \( V_s \) is related to the maximum kinetic energy:
\[ K.E. = e \cdot V_s \]
where \( e \) is the charge of an electron (1 eV). For the first case:
\[ K.E. = 2.4 \text{ eV} \]
Thus, we can write:
\[ 2.0667 - \phi = 2.4 \]
Rearranging gives:
\[ \phi = 2.0667 - 2.4 = -0.3333 \text{ eV} \]

4. **Calculating Energy for Wavelength 4000 Å**:
Now, we calculate the energy for \( \lambda = 4000 \) Å:
\[ E_2 = \frac{12400}{4000} = 3.1 \text{ eV} \]

5. **Finding New Stopping Potential**:
Using the work function found earlier, we can find the new stopping potential:
\[ K.E. = E_2 - \phi \]
\[ K.E. = 3.1 - (-0.3333) = 3.1 + 0.3333 = 3.4333 \text{ eV} \]
Therefore, the stopping potential \( V_s \) is:
\[ V_s = 3.4333 \text{ volts} \]

### Final Answer:
The stopping potential when light of wavelength 4000 Å is used would be approximately **3.43 volts**.

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