Infrared observations of a star show that it is most intense at
a wavelength of 920 nm (9.20 102 nm). What is
the temperature of the star's surface?
_________________________K
If one star has a temperature of 6300 K and another star has a
temperature of 7800 K, how much more energy per second will the
hotter star radiate from each square meter of its surface?
______________________________times
Ans
1
Temperature = 3150 K
2
= 2.35 times
Explanation
1
Most intense wavelength
Let T be the temperature of the star's surface. Therefore using Wein's displacement law
T = 3150 K
2
The temperature of first star T1 = 6300 K
The temperature of second star T2 = 7800 K
The ratio of energy per second per unit area from hotter star to cooler star is
Therefore energy per second from the hotter surface will 2.35 times more.
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