To solve the problem, we will follow these steps:
### Step 1: Understand the given data
- **Intensity of light (I)** = \(3.01 \times 10^{-3} \, \text{W/m}^2\)
- **Side of the square metal surface (s)** = 1 cm = 0.01 m
- **Work function (φ)** = 2 eV
- **Wavelengths of light**: 400 nm, 550 nm, and 650 nm
### Step 2: Calculate the area of the metal surface
The area \(A\) of the square surface is given by:
\[
A = s^2 = (0.01 \, \text{m})^2 = 0.0001 \, \text{m}^2
\]
### Step 3: Calculate the total power incident on the surface
The total power \(P\) incident on the surface can be calculated using the formula:
\[
P = I \times A = (3.01 \times 10^{-3} \, \text{W/m}^2) \times (0.0001 \, \text{m}^2) = 3.01 \times 10^{-7} \, \text{W}
\]
### Step 4: Calculate the energy of photons for each wavelength
The energy \(E\) of a photon is given by:
\[
E = \frac{hc}{\lambda}
\]
where \(h = 6.626 \times 10^{-34} \, \text{J s}\) and \(c = 3 \times 10^8 \, \text{m/s}\).
#### For 400 nm:
\[
\lambda_1 = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}
\]
\[
E_1 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{400 \times 10^{-9}} = 4.97 \times 10^{-19} \, \text{J}
\]
Convert to eV:
\[
E_1 = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.11 \, \text{eV}
\]
#### For 550 nm:
\[
\lambda_2 = 550 \, \text{nm} = 550 \times 10^{-9} \, \text{m}
\]
\[
E_2 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{550 \times 10^{-9}} = 3.61 \times 10^{-19} \, \text{J}
\]
Convert to eV:
\[
E_2 = \frac{3.61 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.26 \, \text{eV}
\]
#### For 650 nm:
\[
\lambda_3 = 650 \, \text{nm} = 650 \times 10^{-9} \, \text{m}
\]
\[
E_3 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{650 \times 10^{-9}} = 3.06 \times 10^{-19} \, \text{J}
\]
Convert to eV:
\[
E_3 = \frac{3.06 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.91 \, \text{eV}
\]
### Step 5: Determine which wavelengths can emit electrons
- **Threshold energy (φ)** = 2 eV
- **400 nm**: \(E_1 = 3.11 \, \text{eV} > 2 \, \text{eV}\) (emission occurs)
- **550 nm**: \(E_2 = 2.26 \, \text{eV} > 2 \, \text{eV}\) (emission occurs)
- **650 nm**: \(E_3 = 1.91 \, \text{eV} < 2 \, \text{eV}\) (no emission)
### Step 6: Calculate the number of photons for each wavelength
The number of photons \(N\) can be calculated using:
\[
N = \frac{P}{E}
\]
#### For 400 nm:
\[
N_1 = \frac{P}{E_1} = \frac{3.01 \times 10^{-7}}{4.97 \times 10^{-19}} \approx 6.05 \times 10^{11} \, \text{photons/s}
\]
#### For 550 nm:
\[
N_2 = \frac{P}{E_2} = \frac{3.01 \times 10^{-7}}{3.61 \times 10^{-19}} \approx 8.33 \times 10^{11} \, \text{photons/s}
\]
### Step 7: Calculate the total number of electrons emitted
Since each photon emits one electron:
\[
N_{\text{total}} = N_1 + N_2 = 6.05 \times 10^{11} + 8.33 \times 10^{11} \approx 1.44 \times 10^{12} \, \text{electrons/s}
\]
### Final Answer:
The number of electrons emitted per second from the metal surface is approximately \(1.44 \times 10^{12}\).
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