Съдържание

Условие Отговори Решение

IOAA 2014, Теория, окончателен вариант — Задача 6. The effective temperature of a star

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Всички материали от IOAA 2014

8th International Olympiad on Astronomy and Astrophysics — Theoretical Test, 12 short problems (final edition)

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Marking scheme

ALL THE CORRECT SOLUTIONS WHICH ARE DIFFERENT FROM THE AUTHOR’S SOLUTION WILL BE EVALUATED AND MARKED ACORDINGLY TO THE MARKING SCHEME.

Условие

From the radiation emitted by a star, two radiations with wavelength values in a narrow range $\Delta\lambda\ll\lambda$ are studied, i.e. the wavelength have values between $\lambda$ and $\lambda+\Delta\lambda$ . According to Planck's relationship (for an absolute black body), the following relation defines, the energy emitted by star in unit time, through a unit area of its surface, per unit wavelength interval:

r=\frac{2\pi hc^2}{\lambda^5\left(e^{\frac{hc}{k\lambda T}}-1\right)}.

The spectral intensities of the radiation with wavelengths $\lambda_1$ and respectively $\lambda_2$ , both within the range $\Delta\lambda$ measured on Earth are $I_1(\lambda_1)$ and $I_2(\lambda_2)$ respectively.

Find out the relation between wavelength $\lambda_1$ and $\lambda_2$ , if $I_1(\lambda_1)=2I_2(\lambda_2)$ , when $hc\ll\lambda kT$ .

Here: h – Planck’s constant; k – Boltzmann’s constant; c – speed of light in vacuum.

$e^x\approx1+x$ if $x\ll1$ .

Отговори

Решение

Problem 6. Marking scheme The effective temperature of a star

3 points

a. We start from the definition of r:

r = \frac{\Delta E}{\Delta t\cdot S_{stea}\cdot\Delta\lambda} = \frac{\Delta E}{\Delta t\cdot 4\pi R^2\cdot\Delta\lambda}

where R is the radius of the star

r = \frac{2\pi hc^2}{\lambda^5\left(e^{hc/k\lambda T} - 1\right)}

Considering d as the distance from the star to the Earth, the definition- relation of the spectral intensity can be written as follows:

I\left(\lambda\right) = \frac{\frac{\Delta E}{\Delta t\cdot\Delta\lambda}}{4\pi d^2}

I\left(\lambda\right) = \frac{2\pi hc^2R^2}{d^2\lambda^5\left(e^{hc/\lambda kT} - 1\right)};

Particularly for each wavelength:

I_1\left(\lambda_1\right) = \frac{2\pi hc^2R^2}{d^2\lambda_1^5\left(e^{hc/\lambda_1kT} - 1\right)};\ I_2\left(\lambda_2\right) = \frac{2\pi hc^2R^2}{d^2\lambda_2^5\left(e^{hc/\lambda_2kT} - 1\right)};

The ratio of the 2 above relations

\frac{I_1\left(\lambda_1\right)}{I_2\left(\lambda_2\right)} = \left(\frac{\lambda_2}{\lambda_1}\right)^5\cdot\frac{e^{hc/\lambda_2kT} - 1}{e^{hc/\lambda_1kT} - 1} \qquad (1)

Represents an equation which allow to find out the temperature of star’s surface T by using spectral measurements

b. If $hc \ll \lambda kT$ , then:

\frac{hc}{k\lambda_1T} \ll 1;\ e^{hc/\lambda_1kT} - 1 \approx 1 + \frac{hc}{k\lambda_1T} - 1 = \frac{hc}{k\lambda_1T}

\frac{hc}{k\lambda_2T} \ll 1;\ e^{hc/\lambda_2kT} - 1 \approx 1 + \frac{hc}{k\lambda_2T} - 1 = \frac{hc}{k\lambda_2T}

The relation (1) becomes:

\frac{I_1\left(\lambda_1\right)}{I_2\left(\lambda_2\right)} = \left(\frac{\lambda_2}{\lambda_1}\right)^5\cdot\frac{\frac{hc}{k\lambda_2T}}{\frac{hc}{k\lambda_2T}} = \left(\frac{\lambda_2}{\lambda_1}\right)^4

I_1 = 2I_2;\ \left(\frac{\lambda_2}{\lambda_1}\right)^4 = 2;\ \lambda_2 = \lambda_1\cdot\sqrt[4]{2} \approx 1{,}2\cdot\lambda_1.

Оригинал в Архива: Short Theoretical.pdf · официални решения: short_final_sol.pdf · допълнителни данни: ioaa1.jpg · допълнителни данни: ioaa2.jpg · допълнителни данни: ioaa3.jpg · допълнителни данни: ioaa4.jpg · допълнителни данни: ioaa5.jpg · допълнителни данни: ioaa6.jpg · допълнителни данни: ioaa7.jpg