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NEET CHEMISTRYMedium

A monochromatic infrared range finder of power 1 milliwatt1 \text{ milliwatt} emits photons with wavelength 1000 nm1000 \text{ nm} in 0.1 second0.1 \text{ second}. The number of photons emitted in 0.1 second0.1 \text{ second} is: (Given: h=6.626×1034 J sh = 6.626 \times 10^{-34} \text{ J s}, c=3×108 m s1c = 3 \times 10^8 \text{ m s}^{-1}, Avogadro number =6.022×1023= 6.022 \times 10^{23})

A

30 \times 10^{37}

B

5 \times 10^{14}

C

30 \times 10^{34}

D

5 \times 10^{11}

Step-by-Step Solution

  1. Calculate the Total Energy Emitted (EtotalE_{\text{total}}): Power (PP) is energy per unit time (P=E/tP = E/t). Given P=1 mW=1×103 J s1P = 1 \text{ mW} = 1 \times 10^{-3} \text{ J s}^{-1} and time t=0.1 st = 0.1 \text{ s}. Etotal=P×t=(1×103 J s1)×(0.1 s)=1×104 JE_{\text{total}} = P \times t = (1 \times 10^{-3} \text{ J s}^{-1}) \times (0.1 \text{ s}) = 1 \times 10^{-4} \text{ J}

  2. Calculate the Energy of One Photon (EphotonE_{\text{photon}}): Using the Planck equation E=hcλE = \frac{hc}{\lambda}: Given λ=1000 nm=1000×109 m=106 m\lambda = 1000 \text{ nm} = 1000 \times 10^{-9} \text{ m} = 10^{-6} \text{ m}. Ephoton=(6.626×1034 J s)(3×108 m s1)106 mE_{\text{photon}} = \frac{(6.626 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m s}^{-1})}{10^{-6} \text{ m}} Ephoton=19.878×1026106 J1.988×1019 JE_{\text{photon}} = \frac{19.878 \times 10^{-26}}{10^{-6}} \text{ J} \approx 1.988 \times 10^{-19} \text{ J}

  3. Calculate the Number of Photons (NN): N=EtotalEphoton=1×104 J1.988×1019 JN = \frac{E_{\text{total}}}{E_{\text{photon}}} = \frac{1 \times 10^{-4} \text{ J}}{1.988 \times 10^{-19} \text{ J}} N0.503×1015=5.03×1014N \approx 0.503 \times 10^{15} = 5.03 \times 10^{14}

The value is closest to 5×10145 \times 10^{14}.

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