To solve the problem step by step, we will follow the outlined process:
### Step 1: Calculate the total number of photons (n)
We know the total energy (E) falling on the surface of the silver sphere is given as \(1 \times 10^{-7} \, J\) and the wavelength (\(\lambda\)) of the UV light is \(200 \, nm\) (which is \(200 \times 10^{-9} \, m\)). The energy of a single photon can be calculated using the formula:
\[
E_{\text{photon}} = \frac{hc}{\lambda}
\]
Where:
- \(h = 6.63 \times 10^{-34} \, J \cdot s\) (Planck's constant)
- \(c = 3 \times 10^{8} \, m/s\) (speed of light)
Now, substituting the values:
\[
E_{\text{photon}} = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^{8})}{200 \times 10^{-9}}
\]
Calculating \(hc\):
\[
hc \approx 1.986 \times 10^{-25} \, J \cdot m
\]
Now substituting this back into the energy of a photon:
\[
E_{\text{photon}} = \frac{1.986 \times 10^{-25}}{200 \times 10^{-9}} = 9.93 \times 10^{-19} \, J
\]
Now, we can find the total number of photons (n):
\[
n = \frac{E}{E_{\text{photon}}} = \frac{1 \times 10^{-7}}{9.93 \times 10^{-19}} \approx 1.01 \times 10^{11} \text{ photons}
\]
### Step 2: Calculate the number of electrons ejected
Given that only 1 out of 1000 photons ejects an electron, the number of electrons ejected (\(n_1\)) can be calculated as:
\[
n_1 = \frac{n}{1000} = \frac{1.01 \times 10^{11}}{1000} = 1.01 \times 10^{8} \text{ electrons}
\]
### Step 3: Calculate the total charge on the sphere
The charge of one electron is approximately \(1.6 \times 10^{-19} \, C\). Therefore, the total charge (\(Q\)) on the sphere can be calculated as:
\[
Q = n_1 \times e = (1.01 \times 10^{8}) \times (1.6 \times 10^{-19}) \approx 1.616 \times 10^{-11} \, C
\]
### Step 4: Calculate the potential on the sphere
The potential (\(V\)) on the surface of a charged sphere is given by the formula:
\[
V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{R}
\]
Where:
- \(\epsilon_0 \approx 8.85 \times 10^{-12} \, C^2/(N \cdot m^2)\)
- \(R = 4.8 \, cm = 0.048 \, m\)
Calculating \( \frac{1}{4 \pi \epsilon_0} \):
\[
\frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^{9} \, N \cdot m^2/C^2
\]
Now substituting the values into the potential formula:
\[
V = 9 \times 10^{9} \cdot \frac{1.616 \times 10^{-11}}{0.048}
\]
Calculating:
\[
V \approx 9 \times 10^{9} \cdot 3.36667 \times 10^{-10} \approx 3.03 \, V
\]
### Final Answer
The potential on the sphere is approximately **3 V**.
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