A metal surface of work function `1.07 eV` is irradiated with light of wavelength `332 nm`. The retarding potential required to stop the escape of photo - electrons is

Options

A. `4.81 eV`
B. `3.74 eV`
C. `2.66 eV`
D. `1.07 eV`

Explanation

To solve the problem, we need to determine the retarding potential required to stop the escape of photoelectrons when a metal surface with a work function of 1.07 eV is irradiated with light of wavelength 332 nm.

### Step-by-Step Solution:

**Step 1: Calculate the energy of the incident photons.**

The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength of the light in meters.

First, convert the wavelength from nanometers to meters:
\[ \lambda = 332 \, \text{nm} = 332 \times 10^{-9} \, \text{m} \]

Now substitute the values into the formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{332 \times 10^{-9} \, \text{m}} \]

Calculating this gives:
\[ E \approx \frac{1.9878 \times 10^{-25}}{332 \times 10^{-9}} \approx 5.98 \times 10^{-19} \, \text{J} \]

To convert this energy from joules to electron volts (1 eV = \(1.6 \times 10^{-19} \, \text{J}\)):
\[ E \approx \frac{5.98 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.74 \, \text{eV} \]

**Step 2: Determine the kinetic energy of the emitted photoelectrons.**

The kinetic energy (KE) of the emitted photoelectrons can be found using the equation:
\[ KE = E - \phi \]
where:
- \(\phi\) is the work function of the metal surface (1.07 eV).

Substituting the values we have:
\[ KE = 3.74 \, \text{eV} - 1.07 \, \text{eV} = 2.67 \, \text{eV} \]

**Step 3: Relate the kinetic energy to the retarding potential.**

The retarding potential \(V\) required to stop the photoelectrons is given by:
\[ KE = eV \]
where \(e\) is the charge of an electron (1 eV = \(1.6 \times 10^{-19} \, \text{J}\)). Rearranging gives:
\[ V = \frac{KE}{e} \]

Since \(KE\) is already in eV, we can directly use:
\[ V = KE = 2.67 \, \text{V} \]

### Final Answer:
The retarding potential required to stop the escape of photoelectrons is **2.67 V**.

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