Calculate the value of retarding potential in volt needed to stop the photoelectron ejected from a metal surface of work function 1.29 eV with light frequency of `5.5 xx 10^(14) sec^(-1)` (nearly)

Answer

To solve the problem of calculating the retarding potential needed to stop the photoelectrons ejected from a metal surface, we can follow these steps:

### Step 1: Understand the given parameters
We have:
- Work function (φ) = 1.29 eV
- Frequency of light (ν) = 5.5 × 10^14 sec^(-1)

### Step 2: Use the photoelectric effect formula
The stopping potential (V) can be calculated using the formula:
\[ V = \frac{hν}{e} - φ \]
where:
- \( h \) = Planck's constant = \( 6.625 × 10^{-34} \) J·s
- \( e \) = Elementary charge = \( 1.6 × 10^{-19} \) C

### Step 3: Calculate the energy of the incident photons
The energy of the incident photons can be calculated using:
\[ E = hν \]
Substituting the values:
\[ E = (6.625 × 10^{-34} \text{ J·s})(5.5 × 10^{14} \text{ sec}^{-1}) \]

### Step 4: Convert the energy from Joules to electron volts
1 eV = \( 1.6 × 10^{-19} \) J, so we convert the energy:
\[ E_{eV} = \frac{E}{1.6 × 10^{-19}} \]

### Step 5: Substitute values into the stopping potential formula
Now, substituting the calculated energy in electron volts and the work function into the stopping potential formula:
\[ V = E_{eV} - φ \]

### Step 6: Calculate the stopping potential
Finally, compute the value of V to find the retarding potential needed to stop the photoelectrons.

### Detailed Calculation:
1. Calculate \( E \):
\[ E = (6.625 × 10^{-34})(5.5 × 10^{14}) = 3.644375 × 10^{-19} \text{ J} \]

2. Convert \( E \) to eV:
\[ E_{eV} = \frac{3.644375 × 10^{-19}}{1.6 × 10^{-19}} \approx 2.277736 \text{ eV} \]

3. Now substitute into the stopping potential formula:
\[ V = 2.277736 - 1.29 \approx 0.987736 \text{ V} \approx 1 \text{ V} \]

### Final Answer:
The retarding potential needed to stop the photoelectrons is approximately **1 V**.

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