Physics 1050 Fall 2002



Spring 2004

Dr Mike Fanelli

Solutions to Assigned Problems:

Chapter 4

PROBLEM 4-3: You are asked to determine the wavelength of a photon for three different energies, one in the red portion of the visible spectrum, one in the infrared, and one in the X-ray part of the spectrum. You are given the energy of each photon, in units of eV, electron volts. The focus of this problem is to illustrate the different energies carried by photons in different parts of the spectrum.

The energy of a photon is related to its frequency: E = h ( (, where h is a number, known as Planck’s constant, whose value is given in the text. ( is the frequency of the radiation. These details are discussed on page 92.

A second fact relates wavelength (the quantity you are trying to determine) to the frequency -- frequency ( wavelength = speed of light for EM radiation. To arrive at an answer, first determine the frequency of each photon using the first expression, then determine the wavelength using the second expression.

Convert electron volts to Joules, since this is the unit in which Planck’s constant is expressed. From MP 4-1 and page 92,

1 eV = 1.6 ( 10(19 Joules and h = 6.63 ( 10(34 Joules sec

a) 2 eV = 3.2 ( 10(19 Joules

E = h ( (, therefore, E ( h = (, after dividing both sides of the expression

by h.

3.2 ( 10(19 Joules ( 6.63 ( 10(34 = (, therefore ( = 4.8 ( 1014 hertz

frequency ( wavelength = speed of light, or, ( ( ( ( c

Dividing both sides by the frequency, (, gives ( ( c ((

( = 3 ( 1010 cm/sec ( 4.8 ( 1014 hertz (the speed of light times the frequency)

= 6.25 ( 10-5 cm. Since 1 nanometer = 10-7 cm, converting to nanometers,

= (6.25 ( 10-5 cm ( 10-7 cm/nm) = 625 nanometers

b) E = 0.1 eV = 1.6 ( 10(20 Joules

( = 2.4 ( 1013 hertz

( = 12,500 nanometers or 12.5 microns, since 1 micron = 1000 nanometers

c) E = 5000 eV = 8.0 ( 10(16 Joules

( = 1.2 ( 1018 hertz

( = 2.5 ( 10-8 cm = 0.25 nanometers

PROBLEM 4-10: This problem concerns the Doppler effect and how the

motion of an object affects the spectral features of that object. A distant galaxy is receding from Earth (motion away) at 3000 km/sec. You are asked to find the shift of the Ly( spectral line

The wavelength of the Ly( spectral line = 121.6 nanometers (from MP 4-1).

The galaxy is moving away from us, so the line should be redshifted (meaning that it will have a longer wavelength).

From the formula on page 100,

Apparent wavelength velocity

---------------------------- = 1 + -------------------,

True wavelength speed of light

? 3000 km/sec

--------------------------- = 1 + ------------------------ = 1.01

121.6 nanometers 300,000 km/sec

? = (121.6 nanometers) ( 1.01 = 122.8 nanometers

Taking the difference, 122.8 – 121.6 = 1.2 nanometers. This is the shift

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