Metal surface emits photoelectrons when light of wavelength 400 nm is incident on it. Calculate the work function of the metal if stopping potential is 0.5 V.

Answer

To solve the problem of calculating the work function of a metal given the wavelength of incident light and the stopping potential, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Wavelength of light, \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \)
- Stopping potential, \( V_s = 0.5 \, \text{V} \)

2. **Convert Stopping Potential to Kinetic Energy:**
The kinetic energy (KE) of the emitted photoelectrons can be calculated using the formula:
\[ KE = e \cdot V_s \]
where \( e \) is the charge of an electron, approximately \( 1.6 \times 10^{-19} \, \text{C} \).
Thus,
\[ KE = 1.6 \times 10^{-19} \, \text{C} \times 0.5 \, \text{V} = 0.8 \times 10^{-19} \, \text{J} \]

3. **Calculate Energy of Incident Light:**
The energy of the incident light can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.63 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \]
\[ E = \frac{1.989 \times 10^{-25} \, \text{J m}}{400 \times 10^{-9} \, \text{m}} = 4.9725 \times 10^{-19} \, \text{J} \]

4. **Convert Energy to Electron Volts:**
To convert energy from joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{4.9725 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 3.105 \, \text{eV} \]

5. **Apply the Photoelectric Equation:**
The photoelectric equation relates the kinetic energy of the emitted electrons to the energy of the incident light and the work function \( \phi \):
\[ KE = E - \phi \]
Rearranging gives us:
\[ \phi = E - KE \]
Substituting the values:
\[ \phi = 3.105 \, \text{eV} - 0.5 \, \text{eV} = 2.605 \, \text{eV} \]

6. **Final Answer:**
The work function of the metal is \( \phi \approx 2.605 \, \text{eV} \).