A. How many photons of a tradiation of wavelength `lamda=5xx10^(-7)` m must fall per second on a blackened plate in order to produce a force of `6.62xx10^(-5)`N?

Options

A. `3xx10^(19)`
B. `5xx10^(22)`
C. `2xx10^(22)`
D. `1.67xx10^(18)`

Explanation

To solve the problem step by step, we need to find out how many photons must fall per second on a blackened plate to produce a force of \(6.62 \times 10^{-5}\) N, given that the wavelength (\(\lambda\)) of the radiation is \(5 \times 10^{-7}\) m.

### Step 1: Understand the relationship between force, momentum, and photons
The force exerted by the photons on the plate can be related to the change in momentum. The momentum \(p\) of a single photon is given by the equation:
\[ p = \frac{h}{\lambda} \]
where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)) and \(\lambda\) is the wavelength.

### Step 2: Calculate the momentum of \(n\) photons
If \(n\) photons are falling on the plate, the total momentum \(P\) contributed by these \(n\) photons is:
\[ P = n \cdot \frac{h}{\lambda} \]

### Step 3: Relate force to momentum
The force \(F\) is the rate of change of momentum, which can be expressed as:
\[ F = \frac{dP}{dt} \]
Since the photons are hitting the plate continuously, we can say:
\[ F = \frac{d(n \cdot \frac{h}{\lambda})}{dt} = n \cdot \frac{h}{\lambda} \]
This means that the force can also be expressed as:
\[ F = n \cdot \frac{h}{\lambda} \]

### Step 4: Rearranging the equation to find \(n\)
From the equation above, we can rearrange it to solve for \(n\):
\[ n = \frac{F \cdot \lambda}{h} \]

### Step 5: Substitute the known values
Now, we can substitute the known values into the equation:
- \(F = 6.62 \times 10^{-5} \, \text{N}\)
- \(\lambda = 5 \times 10^{-7} \, \text{m}\)
- \(h = 6.626 \times 10^{-34} \, \text{Js}\)

Substituting these values gives:
\[ n = \frac{(6.62 \times 10^{-5}) \cdot (5 \times 10^{-7})}{6.626 \times 10^{-34}} \]

### Step 6: Calculate \(n\)
Now, perform the calculation:
\[ n = \frac{3.31 \times 10^{-11}}{6.626 \times 10^{-34}} \approx 5 \times 10^{22} \]

### Conclusion
Thus, the number of photons that must fall per second on the blackened plate to produce the specified force is:
\[ n \approx 5 \times 10^{22} \]

### Final Answer
The answer is \(5 \times 10^{22}\) photons.

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