he product of the number of photons per second and the momentum lost by each photon.

Here’s a step-by-step explanation:

Each photon of wavelength $$\lambda$$ has an energy:

$$E = \frac{hc}{\lambda},$$

where

$$h = 6.62 \times 10^{-34}~\text{J·s}$$ (Planck's constant),

$$c \approx 3 \times 10^{8}~\text{m/s}$$ (speed of light).

The momentum of a photon is given by:

$$p = \frac{E}{c} = \frac{hc}{\lambda c} = \frac{h}{\lambda}.$$

For a completely dark (absorbing) plate, the photon transfers all its momentum upon absorption, so the momentum transfer per photon is $$\frac{h}{\lambda}.$$

If $$n$$ photons hit the plate per second, the total momentum transferred per second (which is the force $$F$$) is:

$$F = n \times \frac{h}{\lambda}.$$

Solving for $$n$$, we have:

$$n = \frac{F \lambda}{h}.$$

Now, plugging in the values:

$$F = 6.62 \times 10^{-5}~\text{N},$$

$$\lambda = 5 \times 10^{-7}~\text{m},$$

$$h = 6.62 \times 10^{-34}~\text{J·s},$$

we get:

$$n = \frac{(6.62 \times 10^{-5}) \times (5 \times 10^{-7})}{6.62 \times 10^{-34}}.$$

Notice that the $$6.62$$ in the numerator and denominator cancel:

$$n = \frac{5 \times 10^{-5-7}}{10^{-34}} = \frac{5 \times 10^{-12}}{10^{-34}}.$$

Dividing the exponents:

$$n = 5 \times 10^{-12 + 34} = 5 \times 10^{22}.$$

Therefore, the number of photons per second $$n$$ is $$5 \times 10^{22}.$$

This corresponds to Option B.

Dual Nature Of Radiation

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