A small `10 W` source of wavelength `99 nm` is held at a distance `0.1 m` from a metal surface. The radius of an atom of the metal is approximately `0.05 nm`. Find (i) the average number of photons striking an atom per second. (ii) the number of photoelectrons emitted per unit area per second if the efficiency of liberation of photoelectrons is `1%`

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (i): Average number of photons striking an atom per second

1. **Calculate the power emitted by the source**:
The power of the source is given as \( P = 10 \, W \). This means it emits \( 10 \, J/s \).

2. **Calculate the surface area of the sphere at a distance of 0.1 m**:
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4 \pi r^2 \]
Substituting \( r = 0.1 \, m \):
\[ A = 4 \pi (0.1)^2 = 4 \pi (0.01) = 0.04 \pi \, m^2 \]

3. **Calculate the energy absorbed per unit area per unit time**:
The energy absorbed per unit area per unit time \( E/A \) is given by:
\[ \frac{E}{A} = \frac{P}{A} = \frac{10}{0.04 \pi} \, J/m^2/s \]

4. **Calculate the energy absorbed per unit time by an atom**:
The radius of an atom is approximately \( r_{atom} = 0.05 \, nm = 0.05 \times 10^{-9} \, m \). The area of the atom \( A_{atom} \) is:
\[ A_{atom} = \pi r_{atom}^2 = \pi (0.05 \times 10^{-9})^2 = \pi (2.5 \times 10^{-21}) \approx 7.85 \times 10^{-21} \, m^2 \]
The energy absorbed per unit time by the atom \( E_{atom} \) is:
\[ E_{atom} = \left(\frac{P}{A}\right) \times A_{atom} \]

5. **Calculate the energy of each photon**:
The energy of a photon \( E_{photon} \) is given by:
\[ E_{photon} = \frac{hc}{\lambda} \]
Where \( h = 6.63 \times 10^{-34} \, J \cdot s \), \( c = 3 \times 10^8 \, m/s \), and \( \lambda = 99 \times 10^{-9} \, m \):
\[ E_{photon} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{99 \times 10^{-9}} \approx 2.01 \times 10^{-18} \, J \]

6. **Calculate the average number of photons striking the atom per second**:
The average number of photons \( N \) striking the atom per second is given by:
\[ N = \frac{E_{atom}}{E_{photon}} \]

### Part (ii): Number of photoelectrons emitted per unit area per second

1. **Calculate the number of photoelectrons emitted**:
The efficiency of liberation of photoelectrons is given as \( 1\% \) or \( 0.01 \). The number of photoelectrons \( N_{photoelectrons} \) emitted per unit area per second is given by:
\[ N_{photoelectrons} = \left(\frac{E}{A \cdot E_{photon}}\right) \times \text{efficiency} \]

2. **Substituting the values**:
Substitute the values calculated earlier into the equation to find \( N_{photoelectrons} \).

### Final Calculation

- For part (i), after substituting the values, you will find the average number of photons striking an atom per second.
- For part (ii), substitute the values into the equation to find the number of photoelectrons emitted per unit area per second.