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Structure of Atom question

2021 · 27 Aug · Shift 2 · Q19
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Structure of Atom question

2021 · 27 Aug · Shift 2 · Q19

JEE MainChemistryStructure of AtomNumerical+4 / −1
The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x ×\times× 1013. The value of x is ‾\underline{\hspace{2cm}}​. (Nearest integer) (h = 6.63 ×\times× 10 −-− 34 Js, c = 3.00 ×\times× 108 ms −-− 1)
Numerical answer
View written solutionFree

Correct answer: 50

  1. Given data
  • Power of infrared source: P=1 mW=1×10−3 WP = 1\,\text{mW} = 1 \times 10^{-3}\,\text{W}P=1mW=1×10−3W
  • Wavelength: λ=1000 nm=1000×10−9 m=1×10−6 m\lambda = 1000\,\text{nm} = 1000 \times 10^{-9}\,\text{m} = 1 \times 10^{-6}\,\text{m}λ=1000nm=1000×10−9m=1×10−6m
  • Time: t=0.1 st = 0.1\,\text{s}t=0.1s
  • Planck's constant: h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s}h=6.63×10−34J s
  • Speed of light: c=3.00×108 m s−1c = 3.00 \times 10^8\,\text{m s}^{-1}c=3.00×108m s−1
  1. Total energy emitted in 0.10.10.1 s

Using Etotal=PtE_{\text{total}} = PtEtotal​=Pt

we get Etotal=(1×10−3)(0.1)=1×10−4 JE_{\text{total}} = (1 \times 10^{-3})(0.1) = 1 \times 10^{-4}\,\text{J}Etotal​=(1×10−3)(0.1)=1×10−4J

  1. Energy of one photon

For a photon, Ephoton=hcλE_{\text{photon}} = \frac{hc}{\lambda}Ephoton​=λhc​

Substitute the values: Ephoton=(6.63×10−34)(3.00×108)1×10−6E_{\text{photon}} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^8)}{1 \times 10^{-6}}Ephoton​=1×10−6(6.63×10−34)(3.00×108)​

Ephoton=19.89×10−20 J=1.989×10−19 JE_{\text{photon}} = 19.89 \times 10^{-20}\,\text{J} = 1.989 \times 10^{-19}\,\text{J}Ephoton​=19.89×10−20J=1.989×10−19J

  1. Number of photons emitted

N=EtotalEphotonN = \frac{E_{\text{total}}}{E_{\text{photon}}}N=Ephoton​Etotal​​

N=1×10−41.989×10−19N = \frac{1 \times 10^{-4}}{1.989 \times 10^{-19}}N=1.989×10−191×10−4​

N≈5.03×1014N \approx 5.03 \times 10^{14}N≈5.03×1014

  1. Write in the form x×1013x \times 10^{13}x×1013

5.03×1014=50.3×10135.03 \times 10^{14} = 50.3 \times 10^{13}5.03×1014=50.3×1013

So, x≈50.3x \approx 50.3x≈50.3

Nearest integer: x=50x = 50x=50

  1. Comparison with stored answer

Stored correct answer = 505050

My derived answer also gives 505050.

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