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Chapter 5: Problem 27

Black holes are objects whose gravity is so strong that not even an object moving at the speed of light can escape from their surface. Hence, black holes do not themselves emit light. But it is possible to detect radiation from material falling toward a black hole. Calculations suggest that as this matter falls, it is compressed and heated to temperatures around \(10^{6}\) K. Calculate the wavelength of maximum emission for this temperature. In what part of the electromagnetic spectrum does this wavelength lie?

Short Answer

Expert verified
The wavelength of maximum emission for the temperature is \(2.898 \times 10^{-9} m\) or 2.898 nm, and it lies in the X-ray part of the electromagnetic spectrum.

Step by step solution

01

Identify Wien's Law

Wien's law formula is \(\lambda_{max} = \frac{b}{T}\) where \( \lambda_{max} \) is the wavelength of maximum spectral radiance, T is the temperature, and b is Wien's displacement constant equal to \(2.898 \times 10^{-3} m \cdot K\).
02

Plug in the given temperature into Wien's Law

Plugging \( T = 10^{6} K \) into the formula, we get: \(\lambda_{max} = \frac{2.898 \times 10^{-3} m \cdot K}{10^{6} K}\)
03

Calculate wavelength

Solving the equation gives the wavelength: \(\lambda_{max} = 2.898 \times 10^{-9} m\) or \(2.898 \times 10^{-3} mm\). Identify that this is in the nanometer (nm) scale.
04

Map the wavelength to the Electromagnetic Spectrum

Finally, with \(\lambda_{max} = 2.898 nm\), refer to the electromagnetics spectrum ranges. The calculated wavelength falls in the X-ray range, which is from 0.01 to 10 nm.

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