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. 1.07 eV
B. 2.66 eV
C. 3.7 eV
D. 4.81 eV

Answer

To solve the problem, we need to find 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 light.**

The energy of the incident light (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) = 332 nm = \( 332 \times 10^{-9} \) m

Substituting the values:
\[ E = \frac{(4.1357 \times 10^{-15} \text{ eV·s})(3 \times 10^8 \text{ m/s})}{332 \times 10^{-9} \text{ m}} \]

Calculating this gives:
\[ E \approx 3.72 \text{ eV} \]

**Step 2: Use the photoelectric equation to find the retarding potential.**

The photoelectric equation is given by:
\[ E = W + eV_0 \]
where:
- \( W \) is the work function of the metal (1.07 eV)
- \( V_0 \) is the retarding potential we want to find
- \( e \) is the charge of an electron (which we will cancel out)

Rearranging the equation to solve for \( V_0 \):
\[ V_0 = \frac{E - W}{e} \]

Substituting the values:
\[ V_0 = \frac{3.72 \text{ eV} - 1.07 \text{ eV}}{1} \]
\[ V_0 = 2.65 \text{ V} \]

### Final Answer:
The retarding potential required to stop the escape of photoelectrons is **2.65 V**.

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