To solve the problem step by step, we will follow these steps:
### Step 1: Calculate the energy of one photon
The energy of a photon can be calculated using the formula:
\[
E = h \cdot f
\]
where \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)) and \( f \) is the frequency of the light.
Since the frequency is just above the threshold frequency, we can relate the energy of the photon to the work function (\( \phi \)):
\[
E = \phi
\]
Given that the work function is \( 6.25 \, \text{eV} \), we convert it to Joules:
\[
\phi = 6.25 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.0 \times 10^{-18} \, \text{J}
\]
### Step 2: Calculate the number of photons emitted per second
The power of the light source is given as \( P = 200 \, \text{W} \). The number of photons emitted per second (\( N \)) can be calculated using:
\[
N = \frac{P}{E}
\]
Substituting the values:
\[
N = \frac{200 \, \text{W}}{1.0 \times 10^{-18} \, \text{J}} = 2.0 \times 10^{20} \, \text{photons/s}
\]
### Step 3: Calculate the kinetic energy gained by the electrons
When the electrons are accelerated through a potential difference (\( V \)), they gain kinetic energy given by:
\[
KE = e \cdot V
\]
where \( e \) is the charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\)) and \( V = 500 \, \text{V} \):
\[
KE = 1.6 \times 10^{-19} \, \text{C} \cdot 500 \, \text{V} = 8.0 \times 10^{-17} \, \text{J}
\]
### Step 4: Calculate the momentum of the electrons
The momentum (\( p \)) of an electron can be calculated using:
\[
p = \sqrt{2m \cdot KE}
\]
where \( m \) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)):
\[
p = \sqrt{2 \cdot 9.1 \times 10^{-31} \, \text{kg} \cdot 8.0 \times 10^{-17} \, \text{J}} = \sqrt{1.456 \times 10^{-46}} \approx 1.208 \times 10^{-23} \, \text{kg m/s}
\]
### Step 5: Calculate the force on the anode
The force (\( F \)) experienced by the anode due to the impact of the electrons can be calculated using:
\[
F = N \cdot \Delta p
\]
where \( \Delta p \) is the change in momentum per electron, which is equal to the momentum calculated above (since the final momentum is zero after absorption):
\[
F = N \cdot p = 2.0 \times 10^{20} \cdot 1.208 \times 10^{-23} \approx 2.416 \times 10^{-3} \, \text{N}
\]
### Step 6: Convert the force to the required format
To express the force in the form \( n \times 10^{-4} \, \text{N} \):
\[
F \approx 2.416 \times 10^{-3} \, \text{N} = 24.16 \times 10^{-4} \, \text{N}
\]
Thus, \( n \approx 24 \).
### Final Answer
The value of \( n \) is:
\[
\boxed{24}
\]
