Question

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 \(2.9 \times 10^{−3} m^{-K}\) and \(\frac{hc}{e} = 1.24 \times 10^{−6} V^{-m}\)]

List-I		List-II
(P)	2000 K	(1)	The radiation at peak wavelength can lead to emission of photoelectrons 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 the widest central maximum of a single slit diffraction.
(S)	10000 K	(4)	The power emitted per unit area is \(\frac{1}{16}\) of that emitted by a blackbody at temperature 6000 K.
		(5)	The radiation at peak emission wavelength can be used to image human bones.

• A. P\(\rightarrow\)3, Q\(\rightarrow\)5, R \(\rightarrow\)2, S\(\rightarrow\)3
• B. P\(\rightarrow\)3, Q\(\rightarrow\)2, R\(\rightarrow\)4, S\(\rightarrow\)1
• C. P\(\rightarrow\)3, Q\(\rightarrow\)4, R\(\rightarrow\)2, S\(\rightarrow\)1
• D. P\(\rightarrow\)1, Q\(\rightarrow\)2, R\(\rightarrow\)5, S\(\rightarrow\)3

Answer 1

The Correct Option is C

To solve this problem, we need to match the temperature of a black body from List-I with the appropriate statement from List-II using Wien's Law and other given properties of blackbody radiation. Wien's Law is given by:

\(\lambda_{\text{max}} = \frac{b}{T}\), where \(b = 2.9 \times 10^{-3} \, \text{m}\cdot\text{K}\). 

Let's calculate \(\lambda_{\text{max}}\) for each temperature in List-I:

1. 2000 K: \(\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{2000}\)
   \(\lambda_{\text{max}} = 1.45 \times 10^{-6} \, \text{m}\) (Infrared region, used in wide diffraction patterns)
2. 3000 K: \(\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{3000}\)
   \(\lambda_{\text{max}} = 9.67 \times 10^{-7} \, \text{m}\) (Near visible region, corresponds to work function explanations)
3. 5000 K: \(\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{5000}\)
   \(\lambda_{\text{max}} = 5.8 \times 10^{-7} \, \text{m}\) (Visible light, human eye sensitivity peak)
4. 10000 K: \(\lambda_{\text{max}} = \frac{2.9 \times 10^{-3}}{10000}\)
   \(\lambda_{\text{max}} = 2.9 \times 10^{-7} \, \text{m}\) (UV region, photoelectron emission possible)
   The power per unit area comparison with 6000 K is necessary for context.

Next, let's match these with the statements in List-II:

1. (P) 2000 K \(\rightarrow\) (3) The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.
2. (Q) 3000 K \(\rightarrow\) (4) The power emitted per unit area is \(\frac{1}{16}\) of that emitted by a blackbody at temperature 6000 K.
3. (R) 5000 K \(\rightarrow\) (2) The radiation at peak wavelength is visible to the human eye.
4. (S) 10000 K \(\rightarrow\) (1) The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function 4 eV.

Correct Matches: P\(\rightarrow\)3, Q\(\rightarrow\)4, R\(\rightarrow\)2, S\(\rightarrow\)1

Answer 2

\[ \lambda_m T = b \]

For option (P) the temperature is minimum, hence \(\lambda_m\) will be maximum.

\[ \lambda_m = \frac{b}{T} = \frac{2.9 \times 10^{-3}}{2000} = 1.45 \times 10^{-6} \, \text{m} = 1450 \, \text{nm} \]

\[ \sin \theta = \frac{\lambda}{d} \]

\[ 2\theta = \text{width of central maximum} \]

\[\Rightarrow \text{width} \propto \lambda \]

\(\Rightarrow\) maximum width for (P)

\[ P \rightarrow 3 \]

For option (Q), \(T = 3000\)

\[ \lambda_m = \frac{b}{T} = \frac{2.9 \times 10^{-3}}{3000} \]

\[ \lambda_m = 0.96 \times 10^{-6} = 966.6 \, \text{nm} \]

\[ P_{3000} = 6A(3000)^4 \]

\[ P_{6000} = 6A(6000)^4 \]

\[ \frac{P_{3000}}{P_{6000}} = \left( \frac{1}{2} \right)^4 = \frac{1}{16} \]

\[ P_{3000} = \frac{1}{16} P_{6000} \]

\[ Q \rightarrow 4 \]

For (R), \(T = 5000 \, \text{K}\)

\[ \lambda_m = \frac{2.9 \times 10^{-3}}{5 \times 10^{3}} = 0.58 \times 10^{-6} = 580 \, \text{nm} \]

\(\Rightarrow\) Visible to human eyes

\[ R \rightarrow 2 \]

For (S), \(T = 10,000 \, \text{K}\)

\[ \lambda_m T = b \Rightarrow \lambda_m = \frac{2.9 \times 10^{-3}}{10,000} = 290 \, \text{nm} \]

\[ \phi = 4 \, \text{eV} \]

\[ \lambda_m = \frac{1240}{4} = 310 \, \text{nm} \]

To emit photoelectron

\[ \lambda_0 < \lambda_T \]

The correct option is (C): P\(\rightarrow\)3, Q\(\rightarrow\)4, R\(\rightarrow\)2, S\(\rightarrow\)1