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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Dual Nature of Matter

JIPMER 2016 · Physics · Dual Nature of Matter

An isotropic point source emits light with wavelength 500 nm . The...

An isotropic point source emits light with wavelength 500 nm . The radiation power of the source is P=10 WP=10 \mathrm{~W}. Find the number of photons passing through unit area per second at a distance of 3 m from the source.An isotropic point source emits light with wavelength 500 nm . The radiation power of the source is P=10 ~W . Find the

Options

  1. A5.92×1017/m2 s5.92 \times 10^{17} / \mathrm{m}^2 \mathrm{~s}
  2. B2.23×1017/m2 s2.23 \times 10^{17} / \mathrm{m}^2 \mathrm{~s}
  3. C2.23×1018/m2 s2.23 \times 10^{18} / \mathrm{m}^2 \mathrm{~s}
  4. D5.92×1018/m2 s5.92 \times 10^{18} / \mathrm{m}^2 \mathrm{~s}

Correct answer

B. 2.23×1017/m2 s2.23 \times 10^{17} / \mathrm{m}^2 \mathrm{~s}

Step-by-step solution

Given, λ=500 nm=500×10−9 m\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}, As, P=n0hcλn0=Pλhc...(i)\begin{aligned} P & =n_0 \frac{h c}{\lambda} \\ n_0 & =\frac{P \lambda}{h c}...(i)\end{aligned} where, n0n_0 is number of photons per second At distance rr from point source, number of photons/area/time An isotropic point source emits light with wavelength 500 nm . The radiation power of the source is P=10 ~W . Find the n′=n04πr2=Pλhc⋅4πr2[n^{\prime}=\frac{n_0}{4 \pi r^2}=\frac{P \lambda}{h c \cdot 4 \pi r^2} \quad[ from Eq. (i)] =10×500×10−96.6×10−34×3×108×4π(3)2=2.23×1017/m2.s\begin{aligned} & =\frac{10 \times 500 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8 \times 4 \pi(3)^2} \\ & =2.23 \times 10^{17} / \mathrm{m}^2 . \mathrm{s}\end{aligned}
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