Question

How many photons of a radiation of wavelength $\lambda=5 \times 10^{-7} m$ must fall per second on a blackened plate in order to produce a force of $6.62 \times 10^{-5} N$ ?

• A. $3 \times 10^{19}$
• B. $5 \times 10^{22}$
• C. $2 \times 10^{22}$
• D. $1.67 \times 10^{18}$

Answer 1

Answer: (B)

The momentum of photon $=h / \lambda$
If $n$ is the number of photons falling per second on the plate, then total momentum per second of the incident photons is
$p=n \times \frac{h}{\lambda}$
Since the plate is blackened, all photons are absorbed by it.
$\frac{\Delta p}{\Delta t}=n \frac{h}{\lambda}$
Since $F=\frac{\Delta p}{\Delta t}=n \frac{h}{\lambda}$
$\therefore n=\frac{F \lambda}{h}=\frac{6.62 \times 10^{-5} \times 5 \times 10^{-7}}{6.62 \times 10^{-34}}$
$=5 \times 10^{22}$