If a photocell is illuminated with a radiation of `1240 A`, then stopping potential is found to be 8 V. The work function of the emitter and the threshold wavelength are

Options

A. a. 1 eV,`5200A`
B. b. 2 eV,`6200A`
C. c. 3 eV,`7200A`
D. d. 4 eV,`4200A`

Explanation

To solve the problem, we need to find the work function of the emitter and the threshold wavelength based on the given information about the photocell. Here’s a step-by-step solution:

### Step 1: Understanding the Problem
We know that when light of a certain wavelength illuminates a photocell, electrons are emitted. The stopping potential (Vs) is the voltage required to stop the emitted electrons. The stopping potential is given as 8 V, and the wavelength of the radiation is 1240 Å (angstroms).

### Step 2: Convert Wavelength to Nanometers
Since we often work in nanometers in the photoelectric effect, we convert the wavelength from angstroms to nanometers:
\[ 1 \text{ Å} = 0.1 \text{ nm} \]
Thus,
\[ 1240 \text{ Å} = 1240 \times 0.1 = 124 \text{ nm} \]

### Step 3: Calculate the Energy of the Incident Radiation
The energy of the incident radiation (E) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 4.14 \times 10^{-15} \text{ eV·s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \text{ m/s} \)
- \( \lambda \) = \( 1240 \text{ Å} = 1240 \times 10^{-10} \text{ m} \)

Substituting the values:
\[ E = \frac{(4.14 \times 10^{-15} \text{ eV·s}) \times (3 \times 10^8 \text{ m/s})}{1240 \times 10^{-10} \text{ m}} \]
Calculating this gives:
\[ E \approx 10 \text{ eV} \]

### Step 4: Relate Stopping Potential to Kinetic Energy
The kinetic energy (KE) of the emitted electrons can be expressed in terms of the stopping potential:
\[ KE = e \cdot V_s \]
Where:
- \( e \) is the charge of an electron (1 eV).
- \( V_s = 8 \text{ V} \)

Thus,
\[ KE = 1 \cdot 8 = 8 \text{ eV} \]

### Step 5: Use the Photoelectric Equation
According to the photoelectric effect, the kinetic energy of the emitted electrons is given by:
\[ KE = E - W \]
Where:
- \( W \) is the work function.

Substituting the values we have:
\[ 8 \text{ eV} = 10 \text{ eV} - W \]
Rearranging gives:
\[ W = 10 \text{ eV} - 8 \text{ eV} = 2 \text{ eV} \]

### Step 6: Calculate the Threshold Wavelength
The threshold wavelength (\( \lambda_0 \)) can be calculated using:
\[ W = \frac{hc}{\lambda_0} \]
Rearranging gives:
\[ \lambda_0 = \frac{hc}{W} \]
Substituting the values:
\[ \lambda_0 = \frac{(4.14 \times 10^{-15} \text{ eV·s}) \times (3 \times 10^8 \text{ m/s})}{2 \text{ eV}} \]
Calculating this gives:
\[ \lambda_0 \approx 620 \text{ nm} = 6200 \text{ Å} \]

### Final Answers
- Work Function \( W = 2 \text{ eV} \)
- Threshold Wavelength \( \lambda_0 = 6200 \text{ Å} \)

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