Question

The maximum in the energy distribution spectrum of the Sun occurs at 4753 $A^\circ$ and 6050K temperature. What will be the temperature of the star in K whose energy distribution shows a maximum at 9506 $A^\circ$ ?

Answer 1

The energy distribution spectrum of any body is governed by the Wien’s displacement law. This law is valid for wavelengths corresponding to maximum emissions.

Formula used: $\lambda T = {\text{ constant}}$ where $\lambda$ is the wavelength at maximum energy of the body in consideration and $T$ is the temperature corresponding to it on the energy distribution curve.

Complete step by step solution:
In this question we are provided with the following data for two different bodies:
Wavelength of max. energy of Sun ${\lambda _S} = 4753A^\circ$ .
Temperature at this wavelength ${T_S} = 6050K$ .
Wavelength of max. energy of star $\lambda = 9506A^\circ$ .
Here, ${A^\circ }$ is called Angstrom and we know that $1{A^\circ } = {10^{ - 10}}m$ .
We are required to find the temperature $T$ of the star.
We know that the Wien’s displacement law in equation form is given as:
$\lambda T = {\text{ constant}}$
So, when we use this equation for the Sun and the given star we get,
${\lambda _S}{T_S} = \lambda T$
Where ${\lambda _s}$ is the wavelength of the sun at maximum energy, ${T_s}$ is the temperature at this wavelength, $\lambda$ is the wavelength of the given star at maximum energy and $T$ is the temperature at this wavelength.
Putting the values in this gives us:
$4753 \times 6050 = 9506 \times T$
Moving the unknown $T$ on the LHS and solving we get,
$T = \dfrac{{4753 \times 6050}}{{9506}}$
$\Rightarrow T = 3025K$
As the units of wavelength are same for both the given stars, the units of Temperature would also remain the same i.e. Kelvin.
$\therefore$ Hence, the temperature of the star is 3025 Kelvins.

Note:
Ideally, Wien’s displacement law works for ideal bodies that absorb and emit all frequencies of light. This law implies that the energy distribution curve for all the bodies follows the same basic curve but the peak occurs at different wavelengths for different temperatures.