Q49 / 51 · JEE Advanced 2023 (Paper 1) · Physics

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-K and hce=1.24×10-6 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 116 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→3, Q→5, R→2, S→3
  • B
    P→3, Q→2, R→4, S→1
  • Correct answer
    C
    P→3, Q→4, R→2, S→1
  • D
    P→1, Q→2, R→5, S→3

Solution

Maximum wavelength for each case:

P 2000 K

λmT=b

⇒λm=bT=2.9×10-32000=1.45×10-6 m=1450 nm

(Q) 3000 K λmT=b⇒λm=2.9×10-33000=966.66 nm
(R) 5000 K λmT=b⇒λm=2.9×10-35000=580 nm
(S) 10000 K λmT=b⇒λm=2.9×10-310,000=290 nm

Now for List-II:

1 λth=hcϕ=1.24×10-64=0.31×10-6 m=310 nm

For λ≤λth, emission of photoelectrons will take place. Therefore, S→1.

2 Wavelength range visible to human eye ranges from 400 to 700 nm. Therefore, R→2.

3 Width of the central maxima is directly proportional to the wavelength of the light used. Therefore, for maximum width, maximum wavelength is required. Hence, P→3.

4 Power emitted by blackbody is given by, 

σT14=116σT42⇒T1=12T2=3000 K

Therefore, Q→4.

5 For imaging bones X-rays are used in the wavelength range of 1-10 nm. None of the options in List-I matches with it.

Hence, the correct answer is option: 'C'.

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