Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Dual Nature of Radiation question

2022 · 26 Jul · Shift 1 · Q47
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Dual Nature of Radiation
  5. /2022 · 26 Jul · Shift 1 · Q47

Dual Nature of Radiation question

2022 · 26 Jul · Shift 1 · Q47

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
A parallel beam of light of wavelength 900 nm900 \mathrm{~nm}900 nm and intensity 100 Wm−2100 \,\mathrm{Wm}^{-2}100Wm−2 is incident on a surface perpendicular to the beam. The number of photons crossing 1 cm21 \mathrm{~cm}^{2}1 cm2 area perpendicular to the beam in one second is :
  1. A
    3×10163 \times 10^{16}3×1016
  2. B
    4.5×10164.5 \times 10^{16}4.5×1016
  3. C
    4.5×10174.5 \times 10^{17}4.5×1017
  4. D
    4.5×10204.5 \times 10^{20}4.5×1020
View written solutionFree

Correct answer: B

  1. Given data
  • Wavelength: λ=900 nm=900×10−9 m\lambda = 900\,\text{nm} = 900 \times 10^{-9}\,\text{m}λ=900nm=900×10−9m
  • Intensity: I=100 W m−2I = 100\,\text{W m}^{-2}I=100W m−2
  • Area: A=1 cm2=10−4 m2A = 1\,\text{cm}^2 = 10^{-4}\,\text{m}^2A=1cm2=10−4m2
  • Time: t=1 st = 1\,\text{s}t=1s
  1. Energy incident on the given area in 1 second

Intensity is energy per unit area per unit time, so

Etotal=IAtE_{\text{total}} = IAtEtotal​=IAt

Substituting,

Etotal=100×10−4×1=10−2 JE_{\text{total}} = 100 \times 10^{-4} \times 1 = 10^{-2}\,\text{J}Etotal​=100×10−4×1=10−2J

  1. Energy of one photon

Using

Eγ=hcλE_{\gamma} = \frac{hc}{\lambda}Eγ​=λhc​

with h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s}h=6.63×10−34J s and c=3×108 m/sc = 3 \times 10^8\,\text{m/s}c=3×108m/s,

Eγ=6.63×10−34×3×108900×10−9E_{\gamma} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{900 \times 10^{-9}}Eγ​=900×10−96.63×10−34×3×108​

Eγ=19.89×10−269×10−7E_{\gamma} = \frac{19.89 \times 10^{-26}}{9 \times 10^{-7}}Eγ​=9×10−719.89×10−26​

Eγ=2.21×10−19 JE_{\gamma} = 2.21 \times 10^{-19}\,\text{J}Eγ​=2.21×10−19J

  1. Number of photons

N=EtotalEγN = \frac{E_{\text{total}}}{E_{\gamma}}N=Eγ​Etotal​​

N=10−22.21×10−19N = \frac{10^{-2}}{2.21 \times 10^{-19}}N=2.21×10−1910−2​

N≈4.52×1016N \approx 4.52 \times 10^{16}N≈4.52×1016

So the number of photons crossing 1 cm21\,\text{cm}^21cm2 area in 1 second is

4.5×1016\boxed{4.5 \times 10^{16}}4.5×1016​

  1. Option check
  • A: 3×10163 \times 10^{16}3×1016 — incorrect
  • B: 4.5×10164.5 \times 10^{16}4.5×1016 — correct
  • C: 4.5×10174.5 \times 10^{17}4.5×1017 — incorrect
  • D: 4.5×10204.5 \times 10^{20}4.5×1020 — incorrect

Therefore, the correct answer is Option B.

PreviousNext

More from Dual Nature of Radiation

  • An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Ee and pe and that of photon are Eph and pph respectively. Which of the following is correct?2022 · MCQ
  • A metal surface is illuminated by a radiation of wavelength 4500 Ao​. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90 ∘ with the magnetic field. If it starts revolving in a…2022 · MCQ
  • The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 Ao​ is 0.42 V. If the threshold frequency is x × 1013 /s, where x is ​ (nearest…2022 · Numerical
  • An electron (mass m) with an initial velocity v=v0​i^(v0​>0) is moving in an electric field E=−E0​i^(E0​>0) where E0​ is constant. If at t=0 de…2022 · MCQ
  • With reference to the observations in photo-electric effect, identify the correct statements from below : (A) The square of maximum velocity of photoelectrons varies linearly with frequency of incident light. (B) The value of saturation…2022 · MCQ
  • An α particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths (λα​:λC12​) is :2022 · MCQ
  • The equation λ=x1.227​ nm can be used to find the de-Brogli wavelength of an electron. In this equation x stands for : Where m= mass of electron P= momentum of electron K=…2022 · MCQ
  • Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively,…2022 · MCQ