Wien Displacement Law MCQ Quiz in বাংলা - Objective Question with Answer for Wien Displacement Law - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Mar 20, 2025
Latest Wien Displacement Law MCQ Objective Questions
Top Wien Displacement Law MCQ Objective Questions
Wien Displacement Law Question 1:
The maximum energy in the thermal radiation from a hot body A occurs at a wavelength of 130 μm. For a second hot body B, the maximum energy in the thermal radiation occurs at a wavelength of 65μm. The relationship between temperature of A (TA) and temperature of B (TB) is -
Answer (Detailed Solution Below)
Wien Displacement Law Question 1 Detailed Solution
Calculation:
We are given that the maximum energy in the thermal radiation from a hot body A occurs at a wavelength of 130 μm. For a second hot body B, the maximum energy in the thermal radiation occurs at a wavelength of 65 μm.
According to Wien's displacement law, the wavelength of the maximum energy emission is inversely proportional to the temperature of the black body, which is given by:
λmax * T = constant
For body A:
λmaxA * TA = constant
For body B:
λmaxB * TB = constant
Since the constant is the same for both bodies, we can write:
λmaxA * TA = λmaxB * TB
Substituting the given wavelengths:
130 μm * TA = 65 μm * TB
Dividing both sides by 65 μm:
2 * TA = TB
Therefore, the relationship between the temperatures is:
Final Answer: TB = 2TA
Wien Displacement Law Question 2:
The spectral energy density uT(λ) vs wavelength (λ) curve of a black body shows a peak at λ = λmax. If the temperature of the black body is doubled, then
Answer (Detailed Solution Below)
Wien Displacement Law Question 2 Detailed Solution
Explanation:
1. Energy Density (u): The formula ( \(u = \sigma T^4 \)) represents the Stefan Boltzmann law, where ( u ) is the energy density, ( \(\sigma\) ) is the Stefan Boltzmann constant, and ( T ) is the temperature. When the temperature is doubled ( \(T \rightarrow 2T \)), the new energy density ( u' ) becomes 16 times the original value, since (\( u' = \sigma (2T)^4 = 16 \sigma T^4\) ).
2. Wien’s Displacement Law (\(λ_m\)): The formula ( \(\lambda_m T = b\) ) represents Wien's displacement law, where ( \(\lambda_m\) ) is the wavelength at the maximum radiation, ( T ) is the temperature, and ( b ) is Wien’s constant. When the temperature doubles ( \(T \rightarrow 2T\) ), the peak wavelength decreases by half ( \(\lambda_m' = \frac{\lambda_m}{2}\) ).
\( \ u = \sigma T^4 \quad \text{as} \quad T \rightarrow 2T \\ u' = \sigma 16T^4 \implies \frac{u'}{u} = 16 \\ \lambda_m T = b \quad \text{as} \quad \lambda_m' 2T = b \\\implies \frac{\lambda_m' 2}{\lambda_m} = 1 \implies \lambda_m' = \frac{\lambda_m}{2} \)
The correct options are (1) and(3) .
Wien Displacement Law Question 3:
A star has a surface temperature of 1000 K. At what wavelength will this star emit most of it’s light? (in nm)
Answer (Detailed Solution Below) 280 - 300
Wien Displacement Law Question 3 Detailed Solution
Explanation:
Using Wien – displacement law,
λmax T = 2.9 × 10-3 mK
\({\lambda _{max}} = \frac{{2.9 \times {{10}^{ - 3}}}}{{1000}}\)
∴ λmax = 2.9 × 10-7 m = 290 nmWien Displacement Law Question 4:
The curve below shows the variation of \({E_{b\lambda }}\) and \(\lambda\) with temperature. Calculate the value of area 1: area 2 + area 1 : area 3 + area 3 : area 2
Answer (Detailed Solution Below) 5.13 - 5.15
Wien Displacement Law Question 4 Detailed Solution
Concept:
Area enclosed by the curve is given by,
\(A = \;\mathop \smallint \nolimits_0^T {E_{b\lambda }} \times d{\rm{\lambda }} = {\rm{\sigma }}{{\rm{T}}^4}\)
Calculation:
A1 = σ (100)4
A2 = σ (200)4
A3 = σ (300)4
A1: A2: A3 = 1: 24: 34
A1: A2 = 1: 24 = 0.0625
A1: A3 = 1: 34 = 0.012
A3: A2 = 34: 24 = 5.06
∴ area 1: area 2 + area 1 : area 3 + area 3 : area 2 = 5.06 + 0.012 + 0.0625
∴ area 1 : area 2 + area 1 : area 3 + area 3 : area 2 = 5.1345
Wien Displacement Law Question 5:
Sun’s surface at 5800 K emits radiation of wavelength 0.5 mm. A furnace of 3000°C will emit radiation of wavelength