Photoelectrons ar ejected from a surface when light of wavelenght `lamda_1=550 nm` is incident on it. The stopping potential for such electrons is `lamda_(S1)=0.19V`. Suppose that radiation of wavelength `lamda_2=190nm` is incident of the surface.
Q. Calulate the work function of the surface.

Options

A. 3.75
B. 2.07
C. 4.2
D. 3.6

Explanation

To calculate the work function of the surface using the given data, we can follow these steps:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect states that when light of sufficient energy strikes a material, it can eject electrons from that material. The energy of the incident light can be expressed in terms of its wavelength.

### Step 2: Use the Energy Equation
The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the light.

### Step 3: Apply the Formula for Stopping Potential
The stopping potential \( V_0 \) relates to the maximum kinetic energy of the emitted photoelectrons:
\[ K.E. = eV_0 \]
where \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

The energy of the incident photon can be expressed as:
\[ E = \phi + K.E. \]
where \( \phi \) is the work function of the material.

### Step 4: Rearranging the Equation
From the above equation, we can express the work function \( \phi \) as:
\[ \phi = E - K.E. \]
Substituting the expressions for energy and kinetic energy gives:
\[ \phi = \frac{hc}{\lambda} - eV_0 \]

### Step 5: Calculate Work Function for \( \lambda_1 = 550 \, \text{nm} \)
1. Convert \( \lambda_1 \) to meters:
\[ \lambda_1 = 550 \, \text{nm} = 550 \times 10^{-9} \, \text{m} \]
2. Calculate the energy of the photon:
\[ E_1 = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{550 \times 10^{-9}} \]
\[ E_1 \approx 3.61 \times 10^{-19} \, \text{J} \]
3. Convert \( V_0 = 0.19 \, \text{V} \) to energy:
\[ K.E. = eV_0 = (1.6 \times 10^{-19})(0.19) \approx 3.04 \times 10^{-20} \, \text{J} \]
4. Now calculate the work function:
\[ \phi = E_1 - K.E. \]
\[ \phi = 3.61 \times 10^{-19} - 3.04 \times 10^{-20} \]
\[ \phi \approx 3.26 \times 10^{-19} \, \text{J} \]

### Step 6: Convert Work Function to Electron Volts
To convert from Joules to electron volts, divide by the charge of an electron:
\[ \phi \approx \frac{3.26 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.04 \, \text{eV} \]

### Step 7: Final Calculation for \( \lambda_2 = 190 \, \text{nm} \)
1. Convert \( \lambda_2 \) to meters:
\[ \lambda_2 = 190 \, \text{nm} = 190 \times 10^{-9} \, \text{m} \]
2. Calculate the energy of the photon:
\[ E_2 = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{190 \times 10^{-9}} \]
\[ E_2 \approx 1.04 \times 10^{-18} \, \text{J} \]
3. The work function remains the same, so:
\[ \phi = E_2 - K.E. \]
\[ \phi = 1.04 \times 10^{-18} - 3.04 \times 10^{-20} \]
\[ \phi \approx 1.01 \times 10^{-18} \, \text{J} \]
4. Convert to electron volts:
\[ \phi \approx \frac{1.01 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 6.31 \, \text{eV} \]

### Final Answer
The work function of the surface is approximately **2.04 eV**.