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Match the temperature of a blackbody listed in Group-I to the corresponding statement in Group-II, and select the correct answer. [Given: Wien’s constant = 2.9 × 10⁻³ m-K and hc/e = 1.24 × 10⁻⁶ V-m] Group-I: (P) 2000 K (Q) 3000 K (R) 5000 K (S) 10000 K Group-II: (1) The peak wavelength of emitted radiation can cause photoelectron ejection from a metal with a work function of 4 eV. (2) The peak wavelength of emitted radiation falls within the visible spectrum. (3) The peak wavelength of emitted radiation produces the broadest central diffraction maximum in a single-slit setup. (4) The energy radiated per unit area is one-sixteenth of that emitted by a blackbody at 6000 K. (5) The peak wavelength of emitted radiation is suitable for imaging human bones.

1. P → 1, Q → 2, R → 5, S → 3
2. P → 3, Q → 2, R → 4, S → 1
3. P → 3, Q → 4, R → 2, S → 1
4. P → 3, Q → 5, R → 2, S → 3

Correct answer: P → 1, Q → 2, R → 5, S → 3

Solution

Using Wien's displacement law, λ_max = b/T, where b is Wien's constant, we calculate the peak wavelength for each temperature. For P (2000 K), the wavelength corresponds to energy sufficient to eject photoelectrons from a metal with a 4 eV work function. For Q (3000 K), the wavelength falls in the visible spectrum. For R (5000 K), the wavelength is suitable for imaging bones. For S (10000 K), the wavelength produces the broadest diffraction maximum.

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