An isotropic point source emits light with wavelength 589 nm. The radiation power of the source is P = 10 W. (a) Find the number of photons passing through unit area per second at a distance of 2 m from the source. (b) Also calculate the distance between the source and the point where the mean concentration of the photons is `100//cm^(3)`.

Answer

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

### Part (a): Finding the number of photons passing through unit area per second at a distance of 2 m from the source.

1. **Given Data:**
- Wavelength, \( \lambda = 589 \, \text{nm} = 589 \times 10^{-9} \, \text{m} \)
- Power of the source, \( P = 10 \, \text{W} \)
- Distance from the source, \( r = 2 \, \text{m} \)

2. **Calculate the energy of a single photon:**
The energy of a single photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) (Planck's constant) is approximately \( 6.626 \times 10^{-34} \, \text{Js} \) and \( c \) (speed of light) is approximately \( 3 \times 10^8 \, \text{m/s} \).

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{589 \times 10^{-9} \, \text{m}} \approx 3.37 \times 10^{-19} \, \text{J} \]

3. **Calculate the total number of photons emitted per second:**
The number of photons emitted per second, \( n_0 \), can be calculated using the formula:
\[ n_0 = \frac{P}{E} \]
Substituting the values:
\[ n_0 = \frac{10 \, \text{W}}{3.37 \times 10^{-19} \, \text{J}} \approx 2.97 \times 10^{19} \, \text{photons/s} \]

4. **Calculate the area of a sphere at distance \( r \):**
The area \( A \) of a sphere is given by:
\[ A = 4\pi r^2 \]
Substituting \( r = 2 \, \text{m} \):
\[ A = 4\pi (2^2) = 16\pi \approx 50.27 \, \text{m}^2 \]

5. **Calculate the number of photons passing through unit area per second:**
The number of photons per unit area per second, \( n \), is given by:
\[ n = \frac{n_0}{A} \]
Substituting the values:
\[ n = \frac{2.97 \times 10^{19} \, \text{photons/s}}{50.27 \, \text{m}^2} \approx 5.91 \times 10^{17} \, \text{photons/m}^2/\text{s} \]

### Part (b): Calculate the distance where the mean concentration of photons is \( 100 \, \text{cm}^{-3} \).

1. **Convert concentration to SI units:**
The mean concentration of photons is given as \( 100 \, \text{cm}^{-3} \), which is:
\[ n = 100 \, \text{cm}^{-3} = 100 \times 10^6 \, \text{m}^{-3} = 10^8 \, \text{m}^{-3} \]

2. **Using the formula for the number of photons in a volume:**
The number of photons per unit volume can be expressed as:
\[ n = \frac{n_0}{4\pi r^2 c} \]
Rearranging for \( r \):
\[ r^2 = \frac{n_0}{4\pi n c} \]
Substituting \( n_0 = 2.97 \times 10^{19} \, \text{photons/s} \), \( n = 10^8 \, \text{m}^{-3} \), and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ r^2 = \frac{2.97 \times 10^{19}}{4\pi (10^8)(3 \times 10^8)} \]
\[ r^2 \approx \frac{2.97 \times 10^{19}}{4 \times 3.14 \times 3 \times 10^{16}} \approx \frac{2.97 \times 10^{19}}{3.77 \times 10^{17}} \approx 78.73 \]
\[ r \approx \sqrt{78.73} \approx 8.88 \, \text{m} \]

### Final Answers:
(a) The number of photons passing through unit area per second at a distance of 2 m from the source is approximately \( 5.91 \times 10^{17} \, \text{photons/m}^2/\text{s} \).

(b) The distance from the source to the point where the mean concentration of photons is \( 100 \, \text{cm}^{-3} \) is approximately \( 8.88 \, \text{m} \).