Category: Astrophysics *

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202207071143 Exercise 18.1 (Q1)

The Sun is a yellow star and emits most of its radiation in the yellow portion of the spectrum. If the sun’s radiation peaks at a frequency of 5.20\times 10^{14}\,\mathrm{Hz}, how much energy is emitted by one photon of this visible yellow light?

J. B., Hickman. (2002). Problem-Solving Exercises in Physics

Roughwork.

In joules,

\begin{aligned} E&=hf \\ & = (6.63\times 10^{-34}\,\mathrm{J\cdot s})(5.20\times 10^{14}\,\mathrm{Hz}) \\ & = 3.4476\times 10^{-19}\,\mathrm{J} \\ \end{aligned}

In electron volts,

\begin{aligned} E & = \frac{3.4476\times 10^{-19}\,\mathrm{J}}{1.60\times 10^{-19}\,\mathrm{J/eV}} \\ & = 2.15475\,\mathrm{eV} \\ \end{aligned}

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202004240535 Homework 1 (Q4)

What is the effective temperature of a neutron star with radius 10^6\,\mathrm{cm} which is detected to emit an energy flux F=9\times 10^{-15}\,\mathrm{erg\, s^{-1}\, cm^{-2}}? The distance to this neutron star is 1\,\mathrm{kpc}.

Setup.

Recall that the effective temperature T_{\textrm{eff}} of a blackbody defines the frequency-integrated intensity of radiation at the source:

\sigma_{\textrm{SB}}T^4_{\textrm{eff}}=F=\displaystyle{\int}I_\nu\cos\theta\,\mathrm{d}\nu\,\mathrm{d}\Omega

where \sigma_{\textrm{SB}}=5.67\times 10^{-5}\,\mathrm{erg\cdot cm^{-2}\, K^{-4}} is the Stefan-Boltzmann constant.

Solution. (guesstimate)

The luminosity L of a source is related to the energy flux F detected at some distance d away from the source by

F=\displaystyle{\frac{L}{4\pi d^2}},

from which one can infer the luminosity of the neutron star in question is

\begin{aligned} L & =9\times 10^{-15}\times 4\pi (3.085677581\times 10^{21})^2 \\ & = 1.076845664\times 10^{30}\enspace \mathrm{erg\, s^{-1}} \end{aligned}

where d=1\,\mathrm{kpc}=3.085677581\times 10^{21}\,\mathrm{cm} is given.

Since L=4\pi R^2\sigma_{\textrm{SB}}T_{\textrm{eff}}^4,

\begin{aligned} T^4_{\textrm{eff}} & =L/4\pi R^2\sigma_{\textrm{SB}} \\ T_{\textrm{eff}} & = \bigg(\frac{L}{4\pi R^2\sigma_{\textrm{SB}}}\bigg)^{1/4} \\ & = \bigg(\frac{1.076845664\times 10^{30}}{4\pi \times (10^6)^2\times 5.67\times 10^{-5}}\bigg)^{1/4} \\ T_{\textrm{eff}} & = 197169.6816\enspace \mathrm{K} \end{aligned}