The photoelectric work function of potassium is 3.0 eV. If light having a wavelength of 2475 Å falls on potassium. Find the stopping potential in volts.

Answer

To solve the problem of finding the stopping potential when light of wavelength 2475 Å falls on potassium with a work function of 3.0 eV, we can follow these steps:

### Step 1: Convert the wavelength from Angstroms to meters
The wavelength is given as 2475 Å. We need to convert this into meters for our calculations.
\[ \text{Wavelength in meters} = 2475 \, \text{Å} = 2475 \times 10^{-10} \, \text{m} \]

### Step 2: Calculate the energy of the incident photons
Using the formula for the energy of a photon:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h = 6.626 \times 10^{-34} \, \text{Js}\) (Planck's constant)
- \(c = 3 \times 10^8 \, \text{m/s}\) (speed of light)
- \(\lambda = 2475 \times 10^{-10} \, \text{m}\)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{2475 \times 10^{-10} \, \text{m}} \]

### Step 3: Calculate the energy in electron volts
Since we need the energy in electron volts (eV), we convert Joules to eV using the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ E \text{ (in eV)} = \frac{E \text{ (in J)}}{1.6 \times 10^{-19}} \]

### Step 4: Use the photoelectric equation
According to Einstein's photoelectric equation:
\[ E = W_0 + KE_{max} \]
Where:
- \(W_0 = 3.0 \, \text{eV}\) (work function)
- \(KE_{max} = eV_0\) (maximum kinetic energy)

Rearranging gives:
\[ KE_{max} = E - W_0 \]

### Step 5: Substitute and solve for stopping potential \(V_0\)
From the equation:
\[ eV_0 = E - W_0 \]
Thus:
\[ V_0 = \frac{E - W_0}{e} \]
Substituting \(W_0 = 3.0 \, \text{eV}\) and the calculated value of \(E\) in eV.

### Step 6: Calculate the stopping potential
After calculating \(E\) and substituting it into the equation, we find \(V_0\).

### Final Calculation:
After performing the calculations, we find:
\[ V_0 = 2.02 \, \text{volts} \]

### Conclusion:
The stopping potential is \(2.02 \, \text{volts}\).

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