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(B) The force exerted by a beam of photons on a perfectly absorbing surface is given by $F = \frac{P}{c}$,where $P$ is the power of the light beam and

Vedclass · Accepted Answer

$05$

Vedclass · Answer

(B) The force exerted by a beam of photons on a perfectly absorbing surface is given by $F = \frac{P}{c}$,where $P$ is the power of the light beam and $c$ is the speed of light.The power $P$ is given by $P = n \cdot E_{photon} = n \cdot \frac{hc}{\lambda}$,where $n$ is the number of photons per second.Substituting $P$ into the force equation: $F = \frac{n \cdot hc}{\lambda \cdot c} = \frac{n \cdot h}{\lambda}$.Rearranging to solve for $n$: $n = \frac{F \cdot \lambda}{h}$.Given $F = 6.62 	imes 10^{-5} \ N$,$\lambda = 5 	imes 10^{-7} \ m$,and $h = 6.62 	imes 10^{-34} \ J \cdot s$.$n = \frac{(6.62 	imes 10^{-5}) 	imes (5 	imes 10^{-7})}{6.62 	imes 10^{-34}}$.$n = \frac{6.62}{6.62} 	imes 5 	imes 10^{-5-7+34} = 1 	imes 5 	imes 10^{22} = 5 	imes 10^{22}$.Thus,$n = 5$.

The number of photons falling per second on a completely darkened plate to produce a force of $6.62 \times 10^{-5} \ N$ is $n$. If the wavelength of the light falling is $5 \times 10^{-7} \ m$,then $n = \_ \times 10^{22}$. (Given: $h = 6.62 \times 10^{-34} \ J \cdot s$)

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