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Dual Nature of Radiation question

2010 · Shift 0 · Q52
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Dual Nature of Radiation question

2010 · Shift 0 · Q52

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
If a source of power 4kW4kW4kW produces 1020{10^{20}}1020 photons/second, the radiation belongs to a part of the spectrum called
  1. A
    XXX -rays
  2. B
    ultraviolet rays
  3. C
    microwaves
  4. D
    γ\gammaγ - rays
View written solutionFree

Correct answer: A

  1. Given data
  • Power of source: P=4 kW=4×103 WP = 4\text{ kW} = 4 \times 10^3\text{ W}P=4 kW=4×103 W
  • Number of photons emitted per second: n=1020 s−1n = 10^{20}\,\text{s}^{-1}n=1020s−1
  1. Energy of one photon

Since power is energy emitted per second,

P=nEP = nEP=nE

So energy per photon,

E=Pn=4×1031020=4×10−17 JE = \frac{P}{n} = \frac{4 \times 10^3}{10^{20}} = 4 \times 10^{-17}\text{ J}E=nP​=10204×103​=4×10−17 J

  1. Convert photon energy to electron volt

Using

1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J}1 eV=1.6×10−19 J

we get

E=4×10−171.6×10−19=2.5×102 eV=250 eVE = \frac{4 \times 10^{-17}}{1.6 \times 10^{-19}} = 2.5 \times 10^2\text{ eV} = 250\text{ eV}E=1.6×10−194×10−17​=2.5×102 eV=250 eV

  1. Identify region of electromagnetic spectrum

A photon energy of about 250 eV250\text{ eV}250 eV lies in the X-ray region.

(Equivalently, wavelength would be around a few nm, also corresponding to soft X-rays.)

  1. Check options
  • A: XXX-rays — Correct
  • B: ultraviolet rays — Too low in energy
  • C: microwaves — Much too low in energy
  • D: γ\gammaγ-rays — Typically much higher in energy

Therefore, the correct answer is A.

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