The energy spectrum of a black body exhibits a maximum around a wavele...
The energy spectrum of a black body exhibits a maximum around a wavele...
Explanation:
The energy spectrum of a black body is described by Planck's law, which states that the energy radiated by a black body at a given wavelength is proportional to the temperature and the intensity of radiation at that wavelength. The intensity of radiation is given by the black body radiation formula:
I(λ, T) = (2hc^2 / λ^5) * (1 / (exp(hc/λkT) - 1))
where I(λ, T) is the intensity of radiation at wavelength λ and temperature T, h is Planck's constant, c is the speed of light, and k is Boltzmann's constant.
Maximum energy wavelength:
The wavelength at which the energy is maximum is given by Wien's displacement law:
λ₀ = b / T
where λ₀ is the maximum energy wavelength, b is Wien's displacement constant (approximately 2.898 x 10^-3 m·K), and T is the temperature.
Change in maximum energy wavelength:
If the temperature is changed such that the energy is maximum around a wavelength of 0.75 λ₀, we can write:
0.75 λ₀ = b / T'
where λ₀ is the original maximum energy wavelength and T' is the new temperature.
Relation between original and new temperature:
Solving the above equation for T', we get:
T' = b / (0.75 λ₀)
Change in power radiated:
The power radiated by the black body is given by Stefan-Boltzmann's law:
P(T) = σ * A * T^4
where P(T) is the power radiated at temperature T, σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W·m^-2·K^-4), and A is the surface area of the black body.
The power radiated at the new temperature T' can be written as:
P(T') = σ * A * T'^4
Substituting the value of T' from above, we get:
P(T') = σ * A * (b / (0.75 λ₀))^4
P(T') = σ * A * (b^4 / (0.75)^4 * λ₀^4)
P(T') = P(T) * (0.75)^-4 * (λ₀)^-4
P(T') = P(T) * (256 / 81)
Therefore, the power radiated by the black body will increase by a factor of 256/81. Hence, the correct answer is option 'D'.
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