When a beam of monochromatic light hits a metal surface , electrons just get ejected with zero kinetic energy . If the intensity of the beam is `79.2xx10^3 W//m^2` and the frequency of light is `2xx10^(15)Hz`, then the number of the photons per metre cube in the radiation is `n xx 10^(13)` . Find the value of n. (Take `h= 6.6xx10^(-34)Js`)

Explanation

To find the number of photons per meter cube in the radiation, we can follow these steps:

### Step 1: Write down the given data
- Intensity \( I = 79.2 \times 10^3 \, \text{W/m}^2 \)
- Frequency \( \nu = 2 \times 10^{15} \, \text{Hz} \)
- Planck's constant \( h = 6.6 \times 10^{-34} \, \text{Js} \)
- Speed of light \( c = 3 \times 10^8 \, \text{m/s} \)

### Step 2: Understand the relationship between intensity, number of photons, and their energy
The intensity \( I \) of the light can be expressed as:
\[ I = n \cdot h \cdot \nu \cdot c \]
where \( n \) is the number of photons per unit volume, \( h \) is Planck's constant, \( \nu \) is the frequency of the light, and \( c \) is the speed of light.

### Step 3: Rearrange the equation to solve for \( n \)
Rearranging the equation gives:
\[ n = \frac{I}{h \cdot \nu \cdot c} \]

### Step 4: Substitute the known values into the equation
Substituting the values we have:
\[ n = \frac{79.2 \times 10^3}{(6.6 \times 10^{-34}) \cdot (2 \times 10^{15}) \cdot (3 \times 10^8)} \]

### Step 5: Calculate the denominator
Calculating the denominator:
\[ h \cdot \nu \cdot c = (6.6 \times 10^{-34}) \cdot (2 \times 10^{15}) \cdot (3 \times 10^8) \]
\[ = 6.6 \times 2 \times 3 \times 10^{-34 + 15 + 8} = 39.6 \times 10^{-11} = 3.96 \times 10^{-10} \]

### Step 6: Calculate \( n \)
Now substituting back into the equation for \( n \):
\[ n = \frac{79.2 \times 10^3}{3.96 \times 10^{-10}} \]
Calculating this gives:
\[ n = 2.0 \times 10^{13} \]

### Final Answer
Thus, the value of \( n \) is:
\[ n = 2 \]