Suggested languages for you:

Americas

Europe

Q11Q

Expert-verifiedFound in: Page 1181

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?**

(a) The infrared lamp emits photons at greater rate.

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

**The expression of emission rate is given by,**

**${\mathit{R}}{\mathbf{=}}\frac{\mathbf{P}}{{\mathbf{E}}_{\mathbf{p}}}$**

**Here, P is power**

**The energy of photon of wavelength ${\mathit{\lambda}}$ is given by,**

role="math" localid="1663136623680" ${{\mathit{E}}}_{{\mathbf{p}}}{\mathbf{=}}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda}}$

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

(a)

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

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

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

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

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

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 photon emitted by the infrared light will be more.

Therefore, the infrared lamp emits photons at greater rate.

(b)

The expression to calculate the emission rate is given by,

*$R=\frac{P\lambda}{hc}$ ** * .... (1)

Substitute 400 W for P, 700 nm for $\lambda $, $6.626\times {10}^{-34}\text{J}\xb7\text{s}$ for h, and $3\times {10}^{8}m{s}^{-1}$ for c in equation (1).

$R=\frac{\left(400W\right)\left(700nm\right)}{\left(6.626\times {10}^{-34}J.s\right)\left(3\times {10}^{8}m/s\right)}\phantom{\rule{0ex}{0ex}}R=\frac{\left(400W\right)\left(700nm\right)\frac{1m}{{10}^{9}nm}}{\left(6.626\times {10}^{-34}J.s\right)\left(3\times {10}^{8}m/s\right)}\phantom{\rule{0ex}{0ex}}R=1.4\times {10}^{21}photon/s$

Therefore, the rate of emitted protons is $1.4\times {10}^{21}photon/s$.

94% of StudySmarter users get better grades.

Sign up for free