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The blackbody at a temperature of 6000K emits a radiation whose intensity spectrum peaks at 600nm. If the temperature is reduced to 300K, the spectrum will peak at
  • a)
    120 μm
  • b)
    12 μm
  • c)
    12 mm
  • d)
    120 mm
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The blackbody at a temperature of 6000K emits a radiation whose intens...
Using Wien's displacement law, we have
λT = Constant
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The blackbody at a temperature of 6000K emits a radiation whose intens...
The peak wavelength of the intensity spectrum emitted by a blackbody is given by Wien's law, which states:

λ_max = b/T

where λ_max is the peak wavelength, T is the temperature in Kelvin, and b is a constant of proportionality equal to approximately 2.898 × 10^-3 m·K.

Using this formula, we can calculate the peak wavelength for the blackbody at 6000K:

λ_max = (2.898 × 10^-3 m·K) / 6000K
λ_max = 4.83 × 10^-7 m
λ_max = 483 nm

Therefore, the peak wavelength of the intensity spectrum emitted by the blackbody at 6000K is 483 nm.

If the temperature is reduced to 300K, we can use Wien's law again to calculate the new peak wavelength:

λ_max = (2.898 × 10^-3 m·K) / 300K
λ_max = 9.66 × 10^-6 m
λ_max = 966 nm

Therefore, the peak wavelength of the intensity spectrum emitted by the blackbody at 300K is 966 nm, which is closer to 1000 nm than to 120 nm.
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