Question

A black body emits radiation at the rate $P$ when its temperature is T. At this temperature the wavelength at which the radiation has maximum intensity is $\lambda_{0} .$ If at another temperature $\mathrm{T}^{\prime \prime}$ the power radiated is $\mathrm{P}^{\prime}$ and wavelength at maximum intensity is $\dfrac{\lambda_{0}}{2}$ then
(A) $P^{\prime} T^{\prime}=32 P T$
(B) $P^{\prime} T^{\prime}=16 P T$
(C) $P^{\prime} T^{\prime}=8 P T$
(D) $P^{\prime} T^{\prime}=4 P T$

Answer 1

We should know that the wavelength is the distance between two wave crests, which is the same as the distance between two troughs. The number of waves that pass-through a given point in one second is called the frequency, measured in units of cycles per second called Hertz. As the full spectrum of visible light travels through a prism, the wavelengths separate into the colours of the rainbow because each colour is a different wavelength. Violet has the shortest wavelength, at around 380 nanometres, and red has the longest wavelength, at around 700 nanometres. Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies. Radio waves, on the other hand, have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation.

Complete step by step answer
From Wien's Displacement Law we know that for a radiating body the product of maximum wavelength radiated $\lambda$ and its temperature T in kelvin is a constant.
$\lambda \times T=\text { Wien's Constant }$
$b=2.898 \times 10^{-3} \mathrm{mK}$
Also, from Stefan-Boltzmann law for radiation from a black body we have Power radiated $P$ is
$P=\varepsilon \sigma A T^{4} \ldots \ldots(2)$
where $\varepsilon$ is emissivity of surface which is $=1$ for a black body, $\sigma$ is Stefan's constant and $A$ is surface area of the radiating object.
At another temperature $T^{\prime}$ we have the expression for the black body
$P^{\prime}=\varepsilon \sigma A\left(T^{\prime}\right)^{4}$Multiplying both sides with $T^{\prime}$, we get
$P^{\prime} T^{\prime}=\varepsilon \sigma A\left(T^{\prime}\right)^{5} \ldots \ldots(5)$
From (1) we have
$\lambda^{\prime} T^{\prime}=b$ $\Rightarrow T^{\prime}=\dfrac{b}{\lambda^{\prime}} \ldots \ldots .(6)$
Inserting given value of wavelength at maximum intensity $\lambda^{\prime}=\dfrac{\lambda}{2}$ in (6) we get
${{T}^{\prime }}=\dfrac{b}{\dfrac{\lambda }{2}}$
$\Rightarrow {{T}^{\prime }}=2\dfrac{b}{\lambda }$
Inserting this in RHS of (5) we get
$P^{\prime} T^{\prime}=\varepsilon \sigma A\left(2 \dfrac{b}{\lambda}\right)^{5}$Rewrite RHS as and then using (1)
${{P}^{\prime }}{{T}^{\prime }}=32\left[ \varepsilon \sigma A{{\left( \dfrac{b}{\lambda } \right)}^{4}} \right]\times \left( \dfrac{b}{\lambda } \right)$
${{P}^{\prime }}{{T}^{\prime }}=32\left[ \varepsilon \sigma A{{T}^{4}} \right]\times T$
Now using (2) we get
$P^{\prime} T^{\prime}=32 P T$

Therefore, the correct answer is Option A.

Note: We should know that black-body radiation has a characteristic, continuous frequency spectrum that depends only on the body's temperature, called the Planck spectrum or Planck's law. As the temperature increases past about 500 degrees Celsius, black bodies start to emit significant amounts of visible light. It occurs due to a process called thermal radiation. Thermal energy causes vibration of molecules or atoms, which in turn vibrates the charge distribution in the material, allowing radiation by the above mechanisms. That radiation, for a perfect absorber, follows the blackbody curve. The primary law governing blackbody radiation is the Planck Radiation Law, which governs the intensity of radiation emitted by unit surface area into a fixed direction from the blackbody as a function of wavelength for a fixed temperature. The mathematical function describing the shape is called the Planck function.