Question

(a) How many photons of a radiation of wavelength $\lambda =5\times {10}^{-7}m$ must fall per second on a blackened plate inorder to produce a force of $6.62\times {10}^{-5}N$?
(b) At what rate will the temperature of plate rise if it's mass is $19.86$ kg and specific heat equal to $2500J\left(kg/K\right)$.

Answer 1

a. 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.
$\therefore \frac{\Delta P}{\Delta t}=n\frac{h}{\lambda }$
Since $F=\frac{\Delta P}{\Delta t}=n\frac{h}{\lambda }\Rightarrow n=\frac{F.\lambda }{h}$
Or $n=\frac{6.62\times {10}^{-5}\times 5\times {10}^{-7}}{6.62\times {10}^{-34}}=5\times {10}^{22}$
b. Energy of each photon $=\frac{hc}{\lambda }$
Since n photons fall on the plate per second, total energy absorbed by the plates in one second is $E=n\times \frac{hc}{\lambda }$
$=1986J{s}^{-1}$
i.e., $\frac{dQ}{dt}=1986J{s}^{-1}$
$mc\frac{dT}{dt}=1986\Rightarrow \frac{dT}{dt}=1986/(19.86\times 2500)$
$=4\times {10}^{-2}{}^{\circ }C{s}^{-1}$