A 100 W sodium lamp $\left(\lambda =589\mathrm{nm}\right)$radiates energy uniformly in all directions. (a) At what rate are photons emitted by the lamp? (b) At what distance from the lamp will a totally absorbing screen absorb photons at the rate of localid="1663064337525" $1.00\mathrm{photon}/{\mathrm{cm}}^2$? (c) What is the photon flux (photons per unit area per unit time) on a small screen 2.00 m from the lamp?

The energy Eof a photon of wavelength $\lambda$is given by,

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

Here, $h$is the Planck’s constant, and $c$is the speed of light.

Assume that the protons are emitted by a rate R from the sodium lamp. Then, the power P of the sodium lamp equals the product of rate R and the energy of each photon E.

$\begin{array}{c}P=RE\\ P=R\left(\frac{hc}{\lambda }\right)\\ R=\frac{P\lambda }{hc}\end{array}$ …… (1)

Substitute 100W for P, 589nm for $\lambda$, localid="1663065939579" $6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}$for h, and localid="1663065946847" $3\times {10}^8\mathrm{m}/\mathrm{s}$for c in equation (1).

Therefore, the rate of emitted protons from the lamp is$2.96\times {10}^{20}\mathrm{photons}/\mathrm{s}.$

The expression for intensity I at a distance r from the lamp is given by,

$\begin{array}{c}I=\frac{R}{4\pi r^2}\\ r=\sqrt{\frac{R}{4\pi I}}......\left(2\right)\end{array}$

$2.96\times {10}^{20}\mathrm{photons}/\mathrm{s}\mathrm{for}\mathrm{R},$and $1\mathrm{photons}/{\mathrm{cm}}^2.\mathrm{s}$for I in equation 2.

Therefore, the required distance from the lamp is$4.86\times {10}^7\mathrm{m}.$

Substitute $2.96\times {10}^{20}\mathrm{photons}/\mathrm{s}$for R, and 2m for r in equation $I=\frac{R}{4\pi r^2}$

$$\begin{gathered} I=\frac{2.96\times {10}^{20}\mathrm{photons}/\mathrm{s}.}{4\pi {\left(2\mathrm{m}\right)}^2} \\ =5.89\times {10}^{18}\mathrm{photons}/{\mathrm{m}}^2.\mathrm{s} \end{gathered}$$

Therefore, the photon flux per unit time is $5.89\times {10}^{18}\mathrm{photons}/{\mathrm{m}}^2.\mathrm{s}$.