The wavelength of photons in two cases are 4000 Å and 3600 Å respectively what is difference in stopping potential for these two ?

Answer

To solve the problem of finding the difference in stopping potential for two wavelengths of photons (4000 Å and 3600 Å), we can follow these steps:

### Step 1: Understand the relationship between energy, wavelength, and stopping potential
The energy of a photon (E) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

The stopping potential (\( V_0 \)) is related to the energy of the photon and the work function (\( \phi \)) of the metal by the equation:
\[ eV_0 = E - \phi \]
where \( e \) is the charge of an electron.

### Step 2: Convert wavelengths from Ångströms to nanometers
1 Å = 0.1 nm, so:
- \( 4000 \, \text{Å} = 400 \, \text{nm} \)
- \( 3600 \, \text{Å} = 360 \, \text{nm} \)

### Step 3: Calculate the energy for each wavelength
**For \( \lambda_1 = 400 \, \text{nm} \):**
\[ E_1 = \frac{hc}{\lambda_1} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \]
Calculating this gives:
\[ E_1 = \frac{1.9878 \times 10^{-25}}{4 \times 10^{-7}} = 4.9695 \times 10^{-19} \, \text{J} \]
To convert this to electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ E_1 = \frac{4.9695 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.105 \, \text{eV} \]

**For \( \lambda_2 = 360 \, \text{nm} \):**
\[ E_2 = \frac{hc}{\lambda_2} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{360 \times 10^{-9} \, \text{m}} \]
Calculating this gives:
\[ E_2 = \frac{1.9878 \times 10^{-25}}{3.6 \times 10^{-7}} = 5.520 \times 10^{-19} \, \text{J} \]
Converting to electron volts:
\[ E_2 = \frac{5.520 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.450 \, \text{eV} \]

### Step 4: Calculate the stopping potential for each case
Using the relation \( eV_0 = E - \phi \):
- For \( V_{0,1} \):
\[ V_{0,1} = E_1 - \phi = 3.105 - \phi \]
- For \( V_{0,2} \):
\[ V_{0,2} = E_2 - \phi = 3.450 - \phi \]

### Step 5: Find the difference in stopping potential
The difference in stopping potential is:
\[ \Delta V_0 = V_{0,2} - V_{0,1} = (3.450 - \phi) - (3.105 - \phi) = 3.450 - 3.105 = 0.345 \, \text{eV} \]

### Final Answer
The difference in stopping potential for the two wavelengths is approximately:
\[ \Delta V_0 \approx 0.345 \, \text{eV} \]