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Thermodynamics question

2023 · 24 Jan · Shift 2 · Q18
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Thermodynamics question

2023 · 24 Jan · Shift 2 · Q18

JEE MainChemistryThermodynamicsNumerical+4 / −1
Following figure shows spectrum of an ideal black body at four different temperatures. The number of correct statement/s from the following is ‾\underline{\hspace{2cm}}​. JEE Main 2023 (Online) 24th January Evening Shift Chemistry - Thermodynamics Question 69 English A. T4>T3>T2>T1\mathrm{T_4 \gt T_3 \gt T_2 \gt T_1}T4​>T3​>T2​>T1​ B. The black body consists of particles performing simple harmonic motion. C. The peak of the spectrum shifts to shorter wavelength as temperature increases. D. T1v1=T2v2=T3v3e{{{T_1}} \over {{v_1}}} = {{{T_2}} \over {{v_2}}} = {{{T_3}} \over {{v_3}}} ev1​T1​​=v2​T2​​=v3​T3​​e constant E. The given spectrum could be explained using quantisation of energy.
Numerical answer
View written solutionFree

Correct answer: 2

  1. Identify the graph behavior for a black body

    For black body radiation:

    • As temperature increases, the total intensity increases.
    • The peak shifts to shorter wavelength.

    So the curve with the peak farthest to the left and highest intensity corresponds to the highest temperature.

  2. Check statement A

    From the spectrum, the leftmost/highest peak corresponds to T4T_4T4​, then T3T_3T3​, then T2T_2T2​, then T1T_1T1​. Hence, T4>T3>T2>T1T_4 > T_3 > T_2 > T_1T4​>T3​>T2​>T1​ So A is correct.

  3. Check statement B

    The classical model that tried to explain black body radiation assumed oscillators behaving like simple harmonic oscillators in the cavity walls. But the statement says:

    "The black body consists of particles performing simple harmonic motion."

    This is not a correct statement about the black body itself. A black body is an ideal absorber/emitter; the radiation explanation involves oscillators in cavity walls, not that the black body "consists of particles performing SHM" as a defining statement.

    So B is incorrect.

  4. Check statement C

    By Wien's displacement law, λmax⁡T=constant\lambda_{\max} T = \text{constant}λmax​T=constant Therefore, when TTT increases, λmax⁡\lambda_{\max}λmax​ decreases.

    So the peak shifts to shorter wavelength as temperature increases. Hence C is correct.

  5. Check statement D

    The statement is T1ν1=T2ν2=T3ν3=constant\frac{T_1}{\nu_1} = \frac{T_2}{\nu_2} = \frac{T_3}{\nu_3} = \text{constant}ν1​T1​​=ν2​T2​​=ν3​T3​​=constant

    Using Wien's law in frequency form, νmax⁡∝T\nu_{\max} \propto Tνmax​∝T Hence, Tνmax⁡=constant\frac{T}{\nu_{\max}} = \text{constant}νmax​T​=constant

    So this relation is correct in principle for the peak frequencies. Therefore D is correct if ν1,ν2,ν3\nu_1,\nu_2,\nu_3ν1​,ν2​,ν3​ denote the corresponding peak frequencies.

    However, in standard black body spectrum graphs shown against wavelength, the marked quantities are often wavelengths, not frequencies. Since the statement explicitly uses ν\nuν, it refers to frequency maxima, and Wien's frequency version indeed gives T/νmax⁡=constantT/\nu_{\max}=\text{constant}T/νmax​=constant.

    Thus D is correct.

  6. Check statement E

    Black body radiation was successfully explained by Planck's quantum theory, i.e. by quantisation of energy.

    So E is correct.

  7. Count correct statements

    Correct statements: A, C, D, E

    Number of correct statements =4= 4=4.

  8. Compare with stored answer

    Stored correct answer is 2, but the derived answer is 4.

    The likely reason for discrepancy is statement D may have been interpreted from a wavelength-spectrum graph while using frequency incorrectly, and statement B might be considered true in a loose textbook sense only for cavity oscillators. Still, as written, the scientifically consistent evaluation gives 4.

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