Question

Black holes in orbit around a normal star are detected from the earth due to the frictional heating of in-falling gas into the black hole, which can reach temperatures greater than ${{10}^{6}}K$. Assuming that the in-falling gas can be modeled as a blackbody radiator, then the wavelength of maximum power lies
A. in the visible region
B. in the X-ray region
C. in the microwave region
D. in the gamma-ray region of electromagnetic spectrum

Answer 1

Hint: As a first step you could recall the expression for energy by Max Planck on blackbody radiation and then express the frequency in terms of wavelength of the gas. Then you could recall the expression for energy based on the kinetic theory of gases. Now you can equate the above two expressions and hence get the required wavelength. Also,
${{k}_{B}}=1.38\times {{10}^{-23}}J{{K}^{-1}}$
Formula used:Planck’s equation,
$E=nh\dfrac{c}{\lambda }$
Expression for energy based on kinetic theory of gases,
$E=\dfrac{3}{2}n{{k}_{B}}T$

Complete answer:
We are given black holes that are orbiting around a normal star. These are detected from the earth due to the frictional heating of in-falling gas into the black hole. Also, the gas can reach a temperature greater than${{10}^{6}}K$. Let us assume that the in-falling gas can be modeled as a blackbody radiator and hence find the wavelength of maximum power.
Based on the Law of distribution of energy in the normal spectrum Max Planck equation is given by,
$E=nh\nu $
Where,$\nu =\dfrac{c}{\lambda }$
Where, c is the universal speed of electromagnetic radiation and λ is its wavelength
Therefore,
$E=nh\dfrac{c}{\lambda }$ ……………………………………. (1)
From kinetic theory of gases we know that the average kinetic energy of a gas is proportional to the absolute temperature and is independent of pressure, volume or nature of the gas.
$E\propto T$
So, for the in falling gas the kinetic energy is given by the expression,
$\Rightarrow E=\dfrac{3}{2}n{{k}_{B}}T$ ………………………………. (2)
Where, the Boltzmann constant given as,
${{k}_{B}}=\dfrac{R}{{{N}_{A}}}=1.38\times {{10}^{-23}}J{{K}^{-1}}$
Energy of the gas given by equations (1) and (2) can now be equated,
$\Rightarrow nh\dfrac{c}{\lambda }=\dfrac{3}{2}n{{k}_{B}}T$
So we get the wavelength as,
$\lambda =\dfrac{2hc}{3{{k}_{B}}T}$
The temperature T of the gas is given as,
$T={{10}^{6}}K$
$\Rightarrow \lambda =\dfrac{2\times 6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{3\times 1.38\times {{10}^{-23}}\times {{10}^{6}}}$
$\Rightarrow \lambda =9.6\times {{10}^{-9}}m$
$\Rightarrow \lambda =9.6nm$
But we know that the x-ray region of the electromagnetic spectrum is marked by the wavelength range of 0.01nm to 10nm. So the wavelength of the given gas lies in between this range. Therefore, the wavelength of maximum power lies in the X-ray region of the electromagnetic spectrum.

So, the correct answer is “Option B”.

Note:
The x-ray region of the electromagnetic spectrum lies in between the gamma ray and ultraviolet region. The UV region lies in the 10nm to 400nm range and the gamma region refers to the region with wavelength less than 0.01nm. Hence, we should be very careful while dealing with the powers of 10.