The threshold frequency for photo electric effect for a metal surface is found to be `4.8xx10^(16)` Hz. The stopping potential required when the metal is irradiated by radiation of frequency `5.6xx10^(16)` Hz is (taking `h=6.6xx10^(-34)Js` and `c=1.6xx10^(-19)C` )

Options

A. 22.4 V
B. 33 V
C. 66 V
D. 198 V

Answer

To solve the problem, we need to find the stopping potential \( V_s \) when a metal surface is irradiated with light of frequency \( f = 5.6 \times 10^{16} \) Hz, given that the threshold frequency \( f_0 = 4.8 \times 10^{16} \) Hz. We will use the photoelectric effect equation:

\[ V_s = \frac{h f - \phi}{e} \]

where:
- \( \phi \) (the work function) can be expressed as \( \phi = h f_0 \)
- \( h \) is Planck's constant
- \( e \) is the charge of an electron

### Step 1: Calculate the Work Function \( \phi \)

First, we calculate the work function \( \phi \):

\[ \phi = h f_0 \]

Substituting the values:

\[ h = 6.6 \times 10^{-34} \, \text{Js} \]
\[ f_0 = 4.8 \times 10^{16} \, \text{Hz} \]

Calculating \( \phi \):

\[ \phi = 6.6 \times 10^{-34} \times 4.8 \times 10^{16} \]

\[ \phi = 3.168 \times 10^{-17} \, \text{J} \]

### Step 2: Calculate the Energy of the Incident Photons

Next, we calculate the energy of the incident photons using the frequency \( f \):

\[ E = h f \]

Substituting the values:

\[ f = 5.6 \times 10^{16} \, \text{Hz} \]

Calculating \( E \):

\[ E = 6.6 \times 10^{-34} \times 5.6 \times 10^{16} \]

\[ E = 3.696 \times 10^{-17} \, \text{J} \]

### Step 3: Calculate the Stopping Potential \( V_s \)

Now, we can substitute \( E \) and \( \phi \) into the stopping potential equation:

\[ V_s = \frac{E - \phi}{e} \]

Substituting the values:

\[ e = 1.6 \times 10^{-19} \, \text{C} \]

Calculating \( V_s \):

\[ V_s = \frac{3.696 \times 10^{-17} - 3.168 \times 10^{-17}}{1.6 \times 10^{-19}} \]

Calculating the numerator:

\[ 3.696 \times 10^{-17} - 3.168 \times 10^{-17} = 0.528 \times 10^{-17} \, \text{J} \]

Now substituting back into the equation for \( V_s \):

\[ V_s = \frac{0.528 \times 10^{-17}}{1.6 \times 10^{-19}} \]

Calculating \( V_s \):

\[ V_s = 3.3 \times 10^{2} \, \text{V} = 3.3 \, \text{V} \]

### Final Answer

The stopping potential required is:

\[ \boxed{3.3 \, \text{V}} \]

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