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Fundamentals Of Physics
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Short Answer

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 1.4×1021 photons/s.

See the step by step solution

Step by Step Solution

Describe the expression of emission rate and energy of the photon

The expression of emission rate is given by,

R=PEp

Here, P is power.

The energy of a photon of wavelength λ is given by,

Ep=hcλ

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.

Number of photons emitted per second=Energy emitted per secondEnergy of each photon

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

Number of photons emitted per second1Energy of each 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=Pλhc…… (1)

Substitute the below values in eq 1.

P=400 wλ=700 nmh=6.626×10-34 J·sc=3×108m/s

R = 400 W700 nm6.626×10-34 J·s3×108 m/s =400 W700 nm1m109nm6.626×10-34 J·s3×108m/s =1.4×1021 photons/s.

Therefore, the rate of emitted protons is 1.4×1021photons/s.

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