JEE Challenger
More from Dual Nature of Radiation and Matter
PhyHardAdv 2023 • Paper 1

Matching Black Body Temperatures with Radiation Characteristics

Match the temperature of a black body given in List-I with an appropriate statement in List-II, and choose the correct option.

[Given: Wien's constant as 2.9×10−3 m-K2.9 \times 10^{-3}\text{ m-K} and hce=1.24×10−6 V-m\frac{hc}{e} = 1.24 \times 10^{-6}\text{ V-m}]

List-IList-II(P) 2000 K(1) The radiation at peak wavelength can lead to emission ofphotoelectrons from a metal of work function 4 eV.(Q) 3000 K(2) The radiation at peak wavelength is visible to human eye.(R) 5000 K(3) The radiation at peak emission wavelength will result in thewidest central maximum of a single slit diffraction.(S) 10000 K(4) The power emitted per unit area is 1/16 of that emitted by ablackbody at temperature 6000 K.(5) The radiation at peak emission wavelength can be used toimage human bones.\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\[8pt] \text{(P) } 2000\text{ K} & \text{(1) The radiation at peak wavelength can lead to emission of} \\ & \quad\text{photoelectrons from a metal of work function } 4\text{ eV.} \\[6pt] \text{(Q) } 3000\text{ K} & \text{(2) The radiation at peak wavelength is visible to human eye.} \\[6pt] \text{(R) } 5000\text{ K} & \text{(3) The radiation at peak emission wavelength will result in the} \\ & \quad\text{widest central maximum of a single slit diffraction.} \\[6pt] \text{(S) } 10000\text{ K} & \text{(4) The power emitted per unit area is } 1/16 \text{ of that emitted by a} \\ & \quad\text{blackbody at temperature } 6000\text{ K.} \\[6pt] & \text{(5) The radiation at peak emission wavelength can be used to} \\ & \quad\text{image human bones.} \end{array}

Options

A

P→3,Q→5,R→2,S→3P \rightarrow 3, Q \rightarrow 5, R \rightarrow 2, S \rightarrow 3

B

P→3,Q→2,R→4,S→1P \rightarrow 3, Q \rightarrow 2, R \rightarrow 4, S \rightarrow 1

C

P→3,Q→4,R→2,S→1P \rightarrow 3, Q \rightarrow 4, R \rightarrow 2, S \rightarrow 1

Correct
D

P→1,Q→2,R→5,S→3P \rightarrow 1, Q \rightarrow 2, R \rightarrow 5, S \rightarrow 3

Topics & Concepts

Dual Nature of Radiation and MatterPhotoelectric EffectDiffraction

Step-by-Step Solution

To determine the correct matching between List-I and List-II, we analyze each temperature given in List-I using the fundamental laws of radiation, diffraction, and the photoelectric effect.

1. Wien's Displacement Law

The peak wavelength λm\lambda_m corresponding to the maximum spectral energy density of a black body at temperature TT is given by Wien's displacement law: λmT=b=2.9×10−3 m⋅K\lambda_m T = b = 2.9 \times 10^{-3} \text{ m}\cdot\text{K}

Let us compute λm\lambda_m for each temperature given in List-I:

  • For (P) TP=2000 KT_P = 2000 \text{ K}: λP=2.9×10−3 m⋅K2000 K=1.45×10−6 m=1450 nm\lambda_P = \frac{2.9 \times 10^{-3} \text{ m}\cdot\text{K}}{2000 \text{ K}} = 1.45 \times 10^{-6} \text{ m} = 1450 \text{ nm}

  • For (Q) TQ=3000 KT_Q = 3000 \text{ K}: λQ=2.9×10−3 m⋅K3000 K≈9.67×10−7 m=967 nm\lambda_Q = \frac{2.9 \times 10^{-3} \text{ m}\cdot\text{K}}{3000 \text{ K}} \approx 9.67 \times 10^{-7} \text{ m} = 967 \text{ nm}

  • For (R) TR=5000 KT_R = 5000 \text{ K}: λR=2.9×10−3 m⋅K5000 K=5.80×10−7 m=580 nm\lambda_R = \frac{2.9 \times 10^{-3} \text{ m}\cdot\text{K}}{5000 \text{ K}} = 5.80 \times 10^{-7} \text{ m} = 580 \text{ nm}

  • For (S) TS=10000 KT_S = 10000 \text{ K}: λS=2.9×10−3 m⋅K10000 K=2.90×10−7 m=290 nm\lambda_S = \frac{2.9 \times 10^{-3} \text{ m}\cdot\text{K}}{10000 \text{ K}} = 2.90 \times 10^{-7} \text{ m} = 290 \text{ nm}

2. Matching Entries in List-I with List-II

Analysis for (P) 2000 K2000 \text{ K}:

  • The width of the central maximum in a single-slit diffraction pattern is given by W=2λDaW = \frac{2\lambda D}{a}, which is directly proportional to the wavelength λ\lambda.
  • Among all four temperatures, TP=2000 KT_P = 2000 \text{ K} produces the maximum peak wavelength (λP=1450 nm\lambda_P = 1450 \text{ nm}).
  • Therefore, the radiation at this peak emission wavelength results in the widest central maximum.
  • P→3P \rightarrow 3

Analysis for (Q) 3000 K3000 \text{ K}:

  • According to the Stefan-Boltzmann law, the total power emitted per unit area of a blackbody is E=σT4E = \sigma T^4.
  • Comparing the emissive power at T=3000 KT = 3000 \text{ K} to that at T0=6000 KT_0 = 6000 \text{ K}: E3000E6000=(30006000)4=(12)4=116\frac{E_{3000}}{E_{6000}} = \left(\frac{3000}{6000}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}
  • Thus, the power emitted per unit area is 116\frac{1}{16} of that emitted at 6000 K6000 \text{ K}.
  • Q→4Q \rightarrow 4

Analysis for (R) 5000 K5000 \text{ K}:

  • The peak wavelength for 5000 K5000 \text{ K} is λR=580 nm\lambda_R = 580 \text{ nm}.
  • This wavelength lies squarely within the human visible spectrum (≈400 nm−700 nm\approx 400 \text{ nm} - 700 \text{ nm}).
  • Thus, the radiation at peak wavelength is visible to the human eye.
  • R→2R \rightarrow 2

Analysis for (S) 10000 K10000 \text{ K}:

  • The peak wavelength for 10000 K10000 \text{ K} is λS=290 nm\lambda_S = 290 \text{ nm}.
  • The photon energy corresponding to this peak wavelength is: E=hcλS=1.24×10−6 eV⋅m2.90×10−7 m≈4.28 eVE = \frac{hc}{\lambda_S} = \frac{1.24 \times 10^{-6} \text{ eV}\cdot\text{m}}{2.90 \times 10^{-7} \text{ m}} \approx 4.28 \text{ eV}
  • Since E=4.28 eV>4.0 eVE = 4.28 \text{ eV} > 4.0 \text{ eV} (the work function of the given metal), the radiation at peak wavelength can cause photoelectric emission.
  • S→1S \rightarrow 1

Conclusion

The correct combination is: P→3,Q→4,R→2,S→1\mathbf{P \rightarrow 3, \quad Q \rightarrow 4, \quad R \rightarrow 2, \quad S \rightarrow 1}

This matches Option C.

PreviousNext