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Dual Nature of Radiation question

2023 · 31 Jan · Shift 1 · Q64
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  5. /2023 · 31 Jan · Shift 1 · Q64

Dual Nature of Radiation question

2023 · 31 Jan · Shift 1 · Q64

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
If a source of electromagnetic radiation having power 15 kW15 \mathrm{~kW}15 kW produces 101610^{16}1016 photons per second, the radiation belongs to a part of spectrum is. (Take Planck constant h=6×10−34Jsh=6 \times 10^{-34} \mathrm{Js}h=6×10−34Js )
  1. A
    Gamma rays
  2. B
    Radio waves
  3. C
    Micro waves
  4. D
    Ultraviolet rays
View written solutionFree

Correct answer: A

  1. Given data

    • Power of source: P=15 kW=1.5×104 WP = 15\text{ kW} = 1.5\times 10^4\text{ W}P=15 kW=1.5×104 W
    • Number of photons emitted per second: n=1016 s−1n = 10^{16}\text{ s}^{-1}n=1016 s−1
    • Planck constant: h=6×10−34 J sh = 6\times 10^{-34}\text{ J s}h=6×10−34 J s

  2. Energy emitted per second Since power is energy per second, total energy emitted each second is: Eper sec=P=1.5×104 JE_{\text{per sec}} = P = 1.5\times 10^4\text{ J}Eper sec​=P=1.5×104 J

  3. Energy of one photon If 101610^{16}1016 photons are emitted per second, then energy of one photon is: E=Pn=1.5×1041016E = \frac{P}{n} = \frac{1.5\times 10^4}{10^{16}}E=nP​=10161.5×104​ E=1.5×10−12 JE = 1.5\times 10^{-12}\text{ J}E=1.5×10−12 J

  4. Find frequency using E=hνE = h\nuE=hν ν=Eh=1.5×10−126×10−34\nu = \frac{E}{h} = \frac{1.5\times 10^{-12}}{6\times 10^{-34}}ν=hE​=6×10−341.5×10−12​ ν=0.25×1022=2.5×1021 Hz\nu = 0.25\times 10^{22} = 2.5\times 10^{21}\text{ Hz}ν=0.25×1022=2.5×1021 Hz

  5. Identify the region of electromagnetic spectrum A frequency of order 1021 Hz10^{21}\text{ Hz}1021 Hz is extremely high and lies in the gamma ray region.

  6. Check options

    • A: Gamma rays ✅
    • B: Radio waves ❌
    • C: Micro waves ❌
    • D: Ultraviolet rays ❌

Therefore, the correct answer is: A: Gamma rays\boxed{\text{A: Gamma rays}}A: Gamma rays​

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