Explanation 1) Concept: Using the formula for the rate at which lamp emits photon, we can find its relation with wavelength. Then comparing the wavelength, we can find the lamp that emits photons at a greater rate and can find its value by plugging values in the above formula. 2) Formulae: i) E = h c λ ii) R = P E 3) Given: i) P u = P i = P = 400 W ii) λ u = 400 n m = 400 × 10 - 9 m iii) λ i = 700 n m = 700 × 10 - 9 m 4) Calculations: a) The energy of photons emitted by the lamp is E = h c λ … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

Fundamentals of Physics, Volume 1, Chapter 1-20

10th Edition

ISBN: 9781118233764

Author: David Halliday

Publisher: WILEY

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Chapter 38, Problem 11P

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To find:

a) The lamp that emits photons at a greater rate

b) The greater rate

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Chapter 38, Problem 10P

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A sodium lamp emits light at the power P = 100 W and at the wavelength = 593 nm, and the emission is 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 1.00 photon /cm²s? (c) What is the rate per square meter at which photons are intercepted by a screen at a distance of 2.40 m from the lamp? (a) Number (b) Number i MI (c) Number i Units Units Units <

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Chapter 38 Solutions

Fundamentals of Physics, Volume 1, Chapter 1-20

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a) The lamp that emits photons at a greater rate b) The greater rate

Solution Summary: The author explains how the infrared lamp emits photons at a greater rate, and calculates its value by plugging values in the above formula.

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To find: a) The lamp that emits photons at a greater rate b) The greater rate

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Chapter 38 Solutions

To find:

a) The lamp that emits photons at a greater rate

b) The greater rate