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. Calculate the stopping potential `V_(S2)`.

Options

A. 4.47
B. 3.16
C. 2.76
D. 5.28

Explanation

To solve the problem, we will use the photoelectric effect equations. The stopping potential is related to the energy of the incident photons and the work function of the material. Let's break down the solution step by step.

### Step 1: Understand the Energy of Incident Photons
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 (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength of the light.

### Step 2: Calculate the Energy for Wavelength \(\lambda_1\)
Given \(\lambda_1 = 550 \, \text{nm} = 550 \times 10^{-9} \, \text{m}\), we can calculate the energy of the photon:
\[ E_1 = \frac{hc}{\lambda_1} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{550 \times 10^{-9} \, \text{m}} \]
Calculating this gives:
\[ E_1 \approx 3.61 \times 10^{-19} \, \text{J} \]

### Step 3: Relate Energy to Stopping Potential
The stopping potential \(V_{S1}\) is related to the energy of the incident photons and the work function \(\phi\):
\[ E_1 = \phi + eV_{S1} \]
where \(e\) is the charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\)). Rearranging gives:
\[ \phi = E_1 - eV_{S1} \]
Substituting \(V_{S1} = 0.19 \, \text{V}\):
\[ \phi = 3.61 \times 10^{-19} \, \text{J} - (1.6 \times 10^{-19} \, \text{C})(0.19 \, \text{V}) \]
Calculating this gives:
\[ \phi \approx 3.61 \times 10^{-19} \, \text{J} - 3.04 \times 10^{-20} \, \text{J} \approx 3.30 \times 10^{-19} \, \text{J} \]

### Step 4: Calculate the Energy for Wavelength \(\lambda_2\)
Now, for \(\lambda_2 = 190 \, \text{nm} = 190 \times 10^{-9} \, \text{m}\):
\[ E_2 = \frac{hc}{\lambda_2} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{190 \times 10^{-9} \, \text{m}} \]
Calculating this gives:
\[ E_2 \approx 1.04 \times 10^{-18} \, \text{J} \]

### Step 5: Relate Energy to Stopping Potential for \(\lambda_2\)
Using the same relation:
\[ E_2 = \phi + eV_{S2} \]
Substituting for \(\phi\):
\[ V_{S2} = \frac{E_2 - \phi}{e} \]
Substituting the values:
\[ V_{S2} = \frac{1.04 \times 10^{-18} \, \text{J} - 3.30 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} \]
Calculating this gives:
\[ V_{S2} \approx \frac{7.10 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} \approx 4.44 \, \text{V} \]

### Final Answer
The stopping potential \(V_{S2}\) for the wavelength \(\lambda_2 = 190 \, \text{nm}\) is approximately:
\[ \boxed{4.44 \, \text{V}} \]