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Question: A hot black body emits the energy at the rate of \(16J{{m}^{-2}}{{s}^{-1}}\)and its most intense rad...

A hot black body emits the energy at the rate of 16Jm−2s−116J{{m}^{-2}}{{s}^{-1}}and its most intense radiation corresponds to 20,000A20,000A. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000A10,000A, then find the value of energy radiated in Jm−2s−1J{{m}^{-2}}{{s}^{-1}},
A.64Jm−2s−1 B.128Jm−2s−1 C.256Jm−2s−1 D.108Jm−2s−1 \begin{aligned} & A.64J{{m}^{-2}}{{s}^{-1}} \\\ & B.128J{{m}^{-2}}{{s}^{-1}} \\\ & C.256J{{m}^{-2}}{{s}^{-1}} \\\ & D.108J{{m}^{-2}}{{s}^{-1}} \\\ \end{aligned}

Explanation

Solution

Wein’s displacement law is the basic concept for solving this question. This is meant that the wavelength of the radiation is inversely proportional to the temperature of the body. Using this law, the temperature is found out which is substituted in the equation of energy. The energy of the radiation is proportional to the fourth power of the temperature. Hope these all will help you to solve this question.

Complete step-by-step answer:
According to the Wein’s displacement law we can write that,
λm∝1T{{\lambda }_{m}}\propto \dfrac{1}{T}
That is we can write that,
λmT=b{{\lambda }_{m}}T=b
Where λm{{\lambda }_{m}} be the maximum wavelength, TTbe the temperature of the body and bbbe a constant known as wein’s constant.
As per the question the maximum wavelength becomes half. Hence according to Wein’s law, the temperature becomes double.
And also we knows that the energy radiation can be written as,
E=σT4E=\sigma {{T}^{4}}
Where σ\sigma be a constant.
Therefore we can write that, the ratio of energy radiated will be,
E1E2=(T1T2)4\dfrac{{{E}_{1}}}{{{E}_{2}}}={{\left( \dfrac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}
As the temperature is doubled in the second case, we can write that the energy of the radiation in the second case will be,
E2=(T2T1)4E1{{E}_{2}}={{\left( \dfrac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}}{{E}_{1}}
It is already mentioned that the energy in the first case is emitted at a rate which is given as,
E1=16Jm−2s−1{{E}_{1}}=16J{{m}^{-2}}{{s}^{-1}}
Substituting this in the equation will give,
E2=(2T1T1)4×16{{E}_{2}}={{\left( \dfrac{2{{T}_{1}}}{{{T}_{1}}} \right)}^{4}}\times 16
Cancelling the common terms and simplifying this will give,
E2=(2)4×16=16×16=256Jm−2s−1{{E}_{2}}={{\left( 2 \right)}^{4}}\times 16=16\times 16=256J{{m}^{-2}}{{s}^{-1}}

So, the correct answer is “Option C”.

Note: Wien's displacement law tells that the curve of a black-body radiation at various temperatures will be maximum at various wavelengths which will be inversely proportional to the temperature of the body. Wien's displacement law is basically the special case of the law by Planck. This law is useful in the case of blackbody radiators.