Question

Black holes are highly condensed remnants of stars. Some black holes, together with a normal star, form binary systems. In such systems the black hole and the normal star orbit about the center of mass of the system. One way black holes can be detected from Earth is by observing the frictional heating of the atmospheric gases from the normal star that fall into the black hole. These gases can reach temperatures greater than 1.10 106 K. Assuming that the falling gas can be modeled as a blackbody radiator, estimate λmax for use in an astronomical detection of a black hole. (Remark: This is in the X-ray region of the electromagnetic spectrum.)

Answer 1

`\lambda_\text{max} = 2.63* 10^(-9)\;\text{m}`.

Step-by-step explanation

The peak emission wavelength of an object depends on its absolute temperature.

`\lambda_\text{max} = (2.90* 10^(-3))/(T)`,

where

• 
  `\lambda_\text{max}` is the wavelength in meters where the emission is the strongest.
• 
  `T` is the temperature of the object in degrees Kelvins.

For the gas falling into the black hole,

`T = 1.10* 10^(6)\;\text{K}`.

Apply the formula:

`\lambda_\text{max} = (2.90* 10^(-3))/(T) = (2.90* 10^(-3))/(1.10* 10^(6)) = 2.64* 10^(-9)\;\text{m} = 2.64 \;\text{nm}`.

The question mentioned that
`\lambda_\text{max}` is in the X-ray region of the electromagnetic spectrum. According to Encyclopedia Britannica, the wavelength of X-rays range from
`10^(-8)\;\text{m}=10\;\text{nm}` to
`10^(-10)\;\text{m} = 0.1\;\text{nm}`, which indeed includes
`2.64* 10^(-9)\;\text{m} = 2.64 \;\text{nm}`.