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Found in: Page 1182

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

A light detector has an absorbing area of ${\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{6}}{\mathbf{}}{{\mathbf{m}}}^{{\mathbf{2}}}$ and absorbs 50% of the incident light, which is at a wavelength 600nm. The detector faces an isotropic source, 12.0 m from the source. The energy E emitted by the source versus time t is given in Fig. 38-26 ( ${{\mathbit{E}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{7}}{\mathbf{.}}{\mathbf{2}}{\mathbit{n}}{\mathbit{J}}$, ${{\mathbit{t}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{2}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{s}}$ ). At what rate are photons absorbed by the detector?

The rate of absorption of the photons by the detector is 6.0 photons/s.

See the step by step solution

Describe the expression of emission rate and energy of the photon:

The expression of emission rate is given by,

${R}{=}\frac{P}{{E}_{p}}$

Here, P is power.

The expression for the frequency of the photon is given by,

${f}{=}\frac{c}{\lambda }$

Here, c is the speed of light and ${\lambda }$ is the wavelength.

The energy of a photon is given by,

E=hf

Here, h is the Planck’s constant.

Combine the above two equations.

${E}{=}\frac{hc}{\lambda }$ ….. (1)

Determine the rate of absorption of the photons by detector:

Consider the given data as below.

The wavelength, $\lambda =600\mathrm{nm}$

Plank’s constant, $h=6.626×{10}^{-34}\mathrm{J}·\mathrm{s}$

The speed of light, $\mathrm{c}=3×{10}^{8}\mathrm{m}/\mathrm{s}$

The time, $\Delta t=2.0\mathrm{s}$

The energy, $\Delta E=7.2\mathrm{nJ}$

Substitute the below values in eq 1.

The wavelength, $\lambda =600\mathrm{nm}$

Plank’s constant, $h=6.626×{10}^{-34}\mathrm{J}·\mathrm{s}$

The speed of light, $c=3×{10}^{8}m/s$

The expression of average power output of the source is given by,

${P}_{\mathrm{emit}}=\frac{\Delta E}{\Delta t}$ ….. (2)

Substitute 7.2 nJ for $\Delta E$, and 2.0 s for $\Delta t$ in equation (2).

Define the rate at which photons radiate.

${R}_{\mathrm{emit}}=\frac{{P}_{\mathrm{emit}}}{E}\phantom{\rule{0ex}{0ex}}=\frac{2.25×{10}^{10}eV/s}{2.07\mathrm{eV}}\phantom{\rule{0ex}{0ex}}=1.09×{10}^{10}photons/s$

Since the source is isotropic, then the flux away is just the power radiated divided by the surface area of the sphere.

$\begin{array}{c}I=\frac{P}{A}\\ =\frac{{P}_{\mathrm{emit}}}{4\pi {r}^{2}}\end{array}$

If the detector absorbs light at 50% efficiency, then the power is,

${P}_{\mathrm{abs}}=\left(\frac{50}{100}\right)I{A}_{\mathrm{detect}}\phantom{\rule{0ex}{0ex}}{\mathrm{R}}_{\mathrm{abs}}\mathrm{E}=\left(0.5\right)\left(\frac{{\mathrm{P}}_{\mathrm{emit}}}{4{\mathrm{\pi r}}^{2}}\right){\mathrm{A}}_{\mathrm{detect}}\phantom{\rule{0ex}{0ex}}{\mathrm{R}}_{\mathrm{abs}}=\left(0.5\right)\left(\frac{{\mathrm{A}}_{\mathrm{detect}}}{4{\mathrm{\pi r}}^{2}}\right)\left(\frac{{\mathrm{P}}_{\mathrm{emit}}}{\mathrm{E}}\right)\phantom{\rule{0ex}{0ex}}$….. (3)

Substitute $2.0×{10}^{-6}{\mathrm{m}}^{2}$ for ${A}_{\mathrm{detect}},$ 12.0m for r, and $1.09×{10}^{10}photons/s$ for ${R}_{\mathrm{emit}}$ in equation (3).

Hence, the rate of absorption of the photons by detector is 6.0 photons/s.