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Expert-verified Found in: Page 1182 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # An ultraviolet lamp emits light of wavelength 400 nm at the rate of 400 W. An infrared lamp emits light of wavelength 700 nm, also at the rate of 400 W. (a) Which lamp emits photons at the greater rate and (b) what is that greater rate?

The infrared lamp emits photons at greater rate.

(b) The rate of emitted protons is $1.4×{10}^{21}\mathrm{photons}/\mathrm{s}.$

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 energy of a photon of wavelength ${\lambda }$ is given by,

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

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

## Determine the lamp that emit photons at the greater rate

(a)

The wavelength of the photon and energy of the photon has an indirect relationship, so; the wavelength of the photon will be larger when the energy is smaller. Here, the wavelength of infrared light is more than ultraviolet light.

Therefore, the energy of the infrared photon is less than the ultraviolet photon. But the power emitted by both lamps is the same.

$\mathrm{Number}\mathrm{of}\mathrm{photons}\mathrm{emitted}\mathrm{per}\mathrm{second}=\frac{\mathrm{Energy}\mathrm{emitted}\mathrm{per}\mathrm{second}}{\mathrm{Energy}\mathrm{of}\mathrm{each}\mathrm{photon}}$

Here, the energy emitted per second is constant and equal to 400 W.

$\mathrm{Number}\mathrm{of}\mathrm{photons}\mathrm{emitted}\mathrm{per}\mathrm{second}\propto \frac{1}{\mathrm{Energy}\mathrm{of}\mathrm{each}\mathrm{photon}}$

From the above relation, it can be observed that if the energy of the infrared photon is less than the ultraviolet photon, then the number of photons emitted by the infrared light will be more.

Therefore, the infrared lamp emits photons at a greater rate.

## Determine the rate of emission of the photon

The expression to calculate the emission rate is given by,

$R=\frac{P\lambda }{hc}$…… (1)

Substitute the below values in eq 1.

$P=400w\phantom{\rule{0ex}{0ex}}\lambda =700nm\phantom{\rule{0ex}{0ex}}h=6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{c}=3×{10}^{8}\mathrm{m}/\mathrm{s}$

Therefore, the rate of emitted protons is $1.4×{10}^{21}\mathrm{photons}/\mathrm{s}.$ ### Want to see more solutions like these? 