Light of wavelength `0.6mum` from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5V. With light of wavelength `0.4mum` from a murcury vapor lamp, the stopping potential is `1.5V`. Then, the work function [in electron volts] of the photocell surface is

Answer

To find the work function (W₀) of the photocell surface, we will use the photoelectric equation:

\[ E - W_0 = eV_s \]

Where:
- \( E \) is the energy of the incident photons,
- \( W_0 \) is the work function,
- \( e \) is the charge of an electron,
- \( V_s \) is the stopping potential.

The energy of the photons can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h \) 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.

### Step 1: Calculate the energy for the sodium lamp (λ = 0.6 μm)

Convert the wavelength from micrometers to meters:

\[ \lambda_1 = 0.6 \, \mu m = 0.6 \times 10^{-6} \, m \]

Now, calculate the energy \( E_1 \):

\[ E_1 = \frac{hc}{\lambda_1} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{0.6 \times 10^{-6} \, m} \]

Calculating this gives:

\[ E_1 = \frac{1.9878 \times 10^{-25}}{0.6 \times 10^{-6}} = 3.313 \times 10^{-19} \, J \]

To convert this energy to electron volts (1 eV = \( 1.6 \times 10^{-19} \, J \)):

\[ E_1 = \frac{3.313 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.07 \, eV \]

### Step 2: Use the stopping potential for sodium lamp

Given the stopping potential \( V_s = 0.5 \, V \):

Using the photoelectric equation:

\[ E_1 - W_0 = eV_s \]

Substituting the values:

\[ 2.07 \, eV - W_0 = 0.5 \, V \]

Rearranging gives:

\[ W_0 = 2.07 \, eV - 0.5 \, V = 1.57 \, eV \]

### Step 3: Calculate the energy for the mercury lamp (λ = 0.4 μm)

Convert the wavelength:

\[ \lambda_2 = 0.4 \, \mu m = 0.4 \times 10^{-6} \, m \]

Calculate the energy \( E_2 \):

\[ E_2 = \frac{hc}{\lambda_2} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{0.4 \times 10^{-6} \, m} \]

Calculating this gives:

\[ E_2 = \frac{1.9878 \times 10^{-25}}{0.4 \times 10^{-6}} = 4.9695 \times 10^{-19} \, J \]

Convert to electron volts:

\[ E_2 = \frac{4.9695 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.10 \, eV \]

### Step 4: Use the stopping potential for mercury lamp

Given the stopping potential \( V_s = 1.5 \, V \):

Using the photoelectric equation:

\[ E_2 - W_0 = eV_s \]

Substituting the values:

\[ 3.10 \, eV - W_0 = 1.5 \, V \]

Rearranging gives:

\[ W_0 = 3.10 \, eV - 1.5 \, V = 1.60 \, eV \]

### Conclusion

The work function \( W_0 \) of the photocell surface is approximately \( 1.60 \, eV \).