PHYS 101
Spring 2009
HW 1 Solutions
Chapter 4
11) a) From Table 4.1: An object at a temperature of 5800 K emits mainly visible
light.
b) Also from Table 4.1: 1 million K is on the dividing line between emitting
mainly ultraviolet light or mainly X-rays. So either ultraviolet or X-rays are
acceptable answers.
c) A person has a temperature of roughly 300 K (98.6 F), so from Table 4.1
primarily emits infrared radiation.
23) Use the inverse square law for light propagation (p. 89): The apparent
brightness of a light source is proportional to the inverse square of the distance of
the light source.
a) Twice as far: 1/22 = 1/4 as bright.
b) Ten times as far: 1/102 = 1/100 as bright.
c) Half as far: 1/(1/2)2 = 4 times as bright.
24) The stars are the same size and distance, the only difference is in their
temperatures, and one star is twice as hot as the other (5800 K and 2900 K). From
the Stefan-Boltzmann law on p. 93, the light output of a star is proportional to its
temperature to the fourth power (T4). The 5800 K star will therefore be 16 times
brighter than the 2900 K star (because 24=16).
26) Use Wien’s Law from p. 93. Solving that equation for T gives T=(3x106)/λmax.
Using λmax=290 nm (given), the star has a temperature of 3x106/290, or 10,300 K.
27) Use the Doppler shift formula on p. 104 to find the velocity: v=c Δλ/λ, where
Δλ is the change in wavelength and is equal to 500.1 nm − 500 nm = 0.1 nm, c is
the speed of light (3x108 m/s), and λ is the wavelength emitted at the star (500
nm). The velocity is then v=(3x108)(0.1)/500, or v=60,000 m/s, or 60 km/s (away
from the Earth because the wavelength increased).