Question

The black body spectrum of an object ${O_1}$ is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of $200{\text{nm}}$. Another object ${O_2}$, has the maximum radiant intensity at $600{\text{nm}}$. The ratio of the power emitted per unit area by source ${O_1}$ to that of source ${O_2}$ is?
(A) $1:81$
(B) $1:9$
(C) $9:1$
(D) $81:1$

Answer 1

Hint: To solve this question, we need to use the Stefan’s law to find the absolute temperature in terms of the wavelength for the two objects. Then these have to be substituted in the Wien’s displacement formula for the radiant power from where the required ratio can be calculated.
Formula used: The formulae used in solving this question are given by
${\lambda _{\max }} = \dfrac{b}{T}$, here ${\lambda _{\max }}$ is the maximum wavelength at the absolute temperature $T$, and $b$ is a constant.
\[{P_R} = \sigma {T^4}\], here \[{P_R}\] is the radiant power at the absolute temperature $T$, and $\sigma $ is the Stefan-Boltzmann constant.

Complete step-by-step solution:
Let the power emitted by the source ${O_1}$ be ${P_1}$ and that emitted by the source ${O_2}$ be ${P_2}$.
We know from the Stefan’s law that the radiant power is proportional to the fourth power of the absolute temperature. So it is given by
\[{P_R} = \sigma {T^4}\]
If the temperature of the source ${O_1}$ is equal to ${T_1}$, then the radiant power emitted by it is given by
${P_1} = \sigma {T_1}^4$........... (1)
Also, if the temperature of the source ${O_2}$ is equal to ${T_2}$, then the radiant power emitted by it is given by
${P_2} = \sigma {T_2}^4$...............(2)
Dividing (2) by (1) we get
$\dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{\sigma {T_2}^4}}{{\sigma {T_1}^4}}$
\[ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = {\left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)^4}\]........(3)
Now, from the Wien’s displacement law we know that the maximum wavelength of the radiation emitted by an object is inversely proportional to its absolute temperature. So it is given by
${\lambda _{\max }} = \dfrac{b}{T}$
$ \Rightarrow T = \dfrac{b}{{{\lambda _{\max }}}}$
If ${\lambda _1}$ is the maximum temperature of the object ${O_1}$ then its absolute temperature is
${T_1} = \dfrac{b}{{{\lambda _1}}}$.........(4)
Also, if the maximum temperature of the object ${O_1}$ then its absolute temperature is
${T_2} = \dfrac{b}{{{\lambda _2}}}$.........(5)
Putting (4) and (5) in (3) we get
\[\dfrac{{{P_2}}}{{{P_1}}} = {\left( {\dfrac{{b/{\lambda _1}}}{{b/{\lambda _2}}}} \right)^4}\]
\[ \Rightarrow \dfrac{{{P_2}}}{{{P_1}}} = {\left( {\dfrac{{{\lambda _2}}}{{{\lambda _1}}}} \right)^4}\]
According to the question, ${\lambda _1} = 200{\text{nm}}$ and ${\lambda _2} = 600{\text{nm}}$. Substituting these above we get
\[\dfrac{{{P_2}}}{{{P_1}}} = {\left( {\dfrac{{600}}{{200}}} \right)^4}\]
On solving we get
\[\dfrac{{{P_2}}}{{{P_1}}} = 81\]

Thus, the ratio of the power emitted per unit area by source ${O_1}$ to that of source ${O_2}$ is equal to $81:1$.

Note: We have been asked to find out the ratio of the power emitted per unit area, but we have evaluated the ratio of power. This is because the two ratios will be equal as the area will get cancelled from both the numerator and the denominator.