The number of photons falling per second on a completely darkened plate to produce a force of `6.62xx10^-5N` is 'n'. If the wavelength of the light falling is `5xx10^-7`m, then n= __________ `xx10^22.`
`(h=6.62 xx10^-34 J-s )`

Options

A. (a)1
B. (b)5
C. (c)0.2
D. (d)3.3

Answer

To solve the problem, we need to find the number of photons falling per second on a completely darkened plate that produces a force of \(6.62 \times 10^{-5} \, \text{N}\) when light of wavelength \(5 \times 10^{-7} \, \text{m}\) is falling on it.

### Step-by-Step Solution:

1. **Understand the relationship between force, momentum, and photons:**
The force \(F\) exerted by the photons can be expressed in terms of the momentum \(P\) of the photons and the number of photons \(n\) falling per second. The momentum of a single photon is given by:
\[ P = \frac{h}{\lambda} \]
where \(h\) is Planck's constant and \(\lambda\) is the wavelength of the light.

2. **Relate force to momentum:**
The force can also be expressed as the rate of change of momentum:
\[ F = \frac{dP}{dt} = n \cdot P \]
Therefore, we can write:
\[ F = n \cdot \frac{h}{\lambda} \]

3. **Rearranging the equation to find \(n\):**
We can rearrange the equation to solve for \(n\):
\[ n = \frac{F \cdot \lambda}{h} \]

4. **Substituting the values:**
Now, substitute the given values into the equation:
- \(F = 6.62 \times 10^{-5} \, \text{N}\)
- \(\lambda = 5 \times 10^{-7} \, \text{m}\)
- \(h = 6.62 \times 10^{-34} \, \text{J-s}\)

Thus, we have:
\[ n = \frac{(6.62 \times 10^{-5}) \cdot (5 \times 10^{-7})}{6.62 \times 10^{-34}} \]

5. **Calculating \(n\):**
- First, calculate the numerator:
\[ 6.62 \times 10^{-5} \cdot 5 \times 10^{-7} = 33.1 \times 10^{-12} = 3.31 \times 10^{-11} \]
- Now, divide by \(h\):
\[ n = \frac{3.31 \times 10^{-11}}{6.62 \times 10^{-34}} = 5 \times 10^{22} \]

6. **Final answer:**
Therefore, the number of photons falling per second, expressed in the form \(n = xx \times 10^{22}\), is:
\[ n = 5 \times 10^{22} \]