Question

A black hole is a blackbody if ever there was one, so it should emit blackbody radiation, called Hawking radiation. A black hole of mass M has a total energy of Mc², a surface area of 161G²M-/c⁴ and a temperature of hc³/16π²kGM. Estimate the typical wavelength of the Hawking radiation emitted by a one-solar- mass (2x10³⁰ kg) black hole. Compare your answer to the size of the black hole.

Answer 1

So, the approximate maximum wavelength of the Hawking radiation emitted by a one-solar-mass black hole is about
`\( 3.69 * 10^(21) \)` meters. This is an extraordinarily large wavelength compared to the size of the black hole, indicating extremely low-energy radiation.

Given:

- Mass of the black hole,
`\( M = 2 * 10^(30) \) kg`

- Planck constant,
`\( h = 6.626 * 10^(-34) \) J*s`

- Speed of light,
`\( c = 3 * 10^8 \) m/s`

- Boltzmann constant,
`\( k = 1.381 * 10^(-23) \) J/K`

- Gravitational constant,
`\( G = 6.674 * 10^(-11) \) N(m/kg)\(^2\)`

The formula for the temperature of a black hole due to Hawking radiation is:

`\[ T = (hc^3)/(16\pi^2kGM) \]`

Let's plug in the values and calculate the temperature:

`\[ T = \frac{(6.626 * 10^(-34) \, \text{J*s})(3 * 10^8 \, \text{m/s})^3}{16\pi^2(1.381 * 10^(-23) \, \text{J/K})(6.674 * 10^(-11) \, \text{N(m/kg)}^2)(2 * 10^(30) \, \text{kg})} \]`

Calculating this:

`\[ T = ((6.626 * 10^(-34))(27 * 10^(24)))/(16\pi^2(1.381 * 10^(-23))(8.890 * 10^(-4))(2 * 10^(30))) \]`

`\[ T = (1.789 * 10^(-9))/(2.199 * 10^4) \]`

`\[ T \approx 8.13 * 10^(-14) \, \text{K} \]`

Now, let's find the maximum wavelength of the Hawking radiation using Wien's displacement law:

`\[ \lambda_{\text{max}} = (c)/(T) \]`

Given:

-
`\( c = 3 * 10^8 \) m/s`

-
`\( T \approx 8.13 * 10^(-14) \) K`

`\[ \lambda_{\text{max}} = (3 * 10^8)/(8.13 * 10^(-14)) \]`

Calculating this:

`\[ \lambda_{\text{max}} \approx 3.69 * 10^(21) \, \text{m} \]`