- #1

jg370

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## Homework Statement

A black hole at 7500K consist of an opening of diameter 0.0500mm, looking into the oven. Find the number of photons per second escaping the hole and having wavelengths between 500nm and 501nm

## Homework Equations

[tex] P = \sigma AeT^4[/tex]

[tex] Watt= \frac{Joule}{sec}[/tex]

[tex] E_{p} = \frac{hc}{\lambda}[/tex]

## The Attempt at a Solution

The total power radiated at the hole is:

[tex] P= 5.67 * 10^{-8} \frac{w}{m^2 k^4}* 1.96*10^{-9} m^2 * (7500K)^4 = 0.35 w[/tex]

I have equated the above answer (0.35 w) according to the following relationship:

[tex] watt = \frac {joule}{sec}[/tex]

Therefore the total energy radiated at the hole is [tex] 0.35 \frac{j}{s}[/tex]

Energy of a photon at the cavity's hole is given by:

[tex] E_p = \frac {hc}{\lambda} = \frac {6.626 *10^{34} js * 3 *10^8 \frac{m}{s}}{500 * 10^{-9}} = 3.97 * 10^{-19} j [/tex]

The number of photons with wavelength 500 nm escaping the hole is:

[tex] N_p = \frac {0.35 \frac{j}{s}}{3.97*10^{-19} j} = \frac{880 *10^{15}}{s}[/tex]

This is not the correct answer; my textbook gives [tex] 1.3*10^{15}[/tex]

I think it my be wrong to equate [tex] w = \frac{joule}{s}[/tex] and I am uncertain what to do about the statement "having wavelengths between 500nm and 501nm" in the question.

Hopefully someone will provide indication(s) that will allow me to get to the correct answer.

Thank you kindly

jg370