To solve the problem, we need to determine the number of electrons emitted per second per unit area from a metal surface when light of a specific wavelength and intensity is incident on it. Here's a step-by-step solution:
### Step 1: Understand the relationship between intensity, number of photons, and energy of photons
The intensity \( I \) of the light is defined as the energy falling per unit area per unit time. The energy of a single photon can be calculated using the formula:
\[
E = \frac{hc}{\lambda}
\]
where:
- \( h \) is Planck's constant (\( 6.64 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the light.
### Step 2: Calculate the energy of a single photon
Given the wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \):
\[
E = \frac{(6.64 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{6000 \times 10^{-10} \, \text{m}}
\]
Calculating this gives:
\[
E = \frac{(6.64 \times 3) \times 10^{-26}}{6000} = \frac{19.92 \times 10^{-26}}{6000} = 3.32 \times 10^{-29} \, \text{J}
\]
### Step 3: Relate intensity to the number of photons
The intensity \( I \) can also be expressed in terms of the number of photons \( N \) incident per unit area per second:
\[
I = \frac{N \cdot E}{A}
\]
where \( A \) is the area (which we can consider as 1 m² for simplicity). Rearranging gives:
\[
N = \frac{I \cdot A}{E}
\]
Substituting \( A = 1 \, \text{m}^2 \):
\[
N = \frac{I}{E}
\]
### Step 4: Substitute the values
Given \( I = 39.6 \, \text{W/m}^2 \):
\[
N = \frac{39.6 \, \text{W/m}^2}{3.32 \times 10^{-29} \, \text{J}} \approx 1.19 \times 10^{30} \, \text{photons/m}^2 \text{s}
\]
### Step 5: Calculate the number of emitted electrons
Since only 1% of the incident photons lead to the emission of electrons:
\[
\text{Electrons emitted} = 0.01 \times N = 0.01 \times 1.19 \times 10^{30} \approx 1.19 \times 10^{28} \, \text{electrons/m}^2 \text{s}
\]
### Step 6: Final result
Thus, the number of electrons emitted per second per unit area from the surface is:
\[
\text{Electrons emitted} = 1.19 \times 10^{28} \, \text{electrons/m}^2 \text{s}
\]
