A black-body at temperature T emits radiation of peak wavelength λ. If the temperature of a body is made double, peak wavelength will shift to:
(A) λ/2
(B) 2λ
(C) λ/4
(D) 4λ
Answer (2)
To solve this problem, we rely on Wien's Displacement Law, which relates the temperature of a black-body to the wavelength at which it emits radiation most strongly. Specifically, Wien's Law states that:
λ ma x = T b
where:
λ ma x is the peak wavelength, T is the absolute temperature of the black-body, and b is Wien's constant, approximately equal to 2.898 × 1 0 − 3 meter kelvin (m·K).
According to Wien’s law, when the temperature T of the body is increased, the peak wavelength λ decreases proportionally.
Given that the temperature T is doubled, the new temperature T ′ = 2 T . Applying Wien’s Displacement Law to both the initial and the new conditions:
λ = T b
and for the new condition:
λ ′ = T ′ b = 2 T b = 2 1 ( T b ) = 2 λ
This means the peak wavelength λ ′ will become half of the original peak wavelength λ when the temperature is doubled.
Therefore, the peak wavelength will shift to λ /2 when the temperature is doubled.
The correct answer is (A) λ /2 .
When the temperature of the black-body is doubled, the peak wavelength of emitted radiation shifts to half of its original value, specifically λ /2 . This conclusion is based on Wien's Displacement Law. Thus, the correct answer is (A) λ /2 .
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