Calculate the change in stopping potential for a photosensitive surface when frequency of light on the surface changes from `2xx10^(15)` Hz to `6xx10^(15)Hz`

Answer

To calculate the change in stopping potential for a photosensitive surface when the frequency of light changes from \(2 \times 10^{15}\) Hz to \(6 \times 10^{15}\) Hz, we can follow these steps:

### Step 1: Understand the Energy of Photons
The energy of a photon is given by the equation:
\[ E = h \cdot \nu \]
where \(E\) is the energy, \(h\) is Planck's constant (\(6.63 \times 10^{-34} \, \text{Js}\)), and \(\nu\) is the frequency of the light.

### Step 2: Calculate Initial and Final Energies
1. **Initial Energy (\(E_i\))** at frequency \(2 \times 10^{15}\) Hz:
\[ E_i = h \cdot (2 \times 10^{15}) = 6.63 \times 10^{-34} \cdot 2 \times 10^{15} \]
\[ E_i = 6.63 \times 2 \times 10^{-19} = 13.26 \times 10^{-19} \, \text{J} \]

2. **Final Energy (\(E_f\))** at frequency \(6 \times 10^{15}\) Hz:
\[ E_f = h \cdot (6 \times 10^{15}) = 6.63 \times 10^{-34} \cdot 6 \times 10^{15} \]
\[ E_f = 6.63 \times 6 \times 10^{-19} = 39.78 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate Change in Energy
The change in energy (\(\Delta E\)) is given by:
\[ \Delta E = E_f - E_i \]
Substituting the values:
\[ \Delta E = (39.78 \times 10^{-19}) - (13.26 \times 10^{-19}) = 26.52 \times 10^{-19} \, \text{J} \]

### Step 4: Convert Energy to Electron Volts
To convert joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ \Delta E \text{ (in eV)} = \frac{26.52 \times 10^{-19}}{1.6 \times 10^{-19}} = 16.575 \, \text{eV} \]

### Step 5: Relate Energy Change to Stopping Potential
The stopping potential (\(V_s\)) is related to the energy of the photons:
\[ \Delta E = e \cdot V_s \]
where \(e\) is the charge of an electron. Thus, the change in stopping potential is:
\[ V_s = \Delta E \]
So, the change in stopping potential is:
\[ V_s = 16.575 \, \text{V} \]

### Final Answer
The change in stopping potential when the frequency of light changes from \(2 \times 10^{15}\) Hz to \(6 \times 10^{15}\) Hz is approximately **16.575 V**.

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