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 $$2500 J (kg/K)$$.

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