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

BITSAT 2016 · Physics · Dual Nature of Matter

Find the number of photon emitted per second by a 25 watt source of...

Find the number of photon emitted per second by a 25 watt source of monochromatic light of wavelength 6600A˚6600 Å. What is the photoelectric current assuming 3%3 \% efficiency for photoelectric effect?

Options

  1. A253×1019J,0.4amp\frac{25}{3} \times 10^{19} \mathrm{ J}, 0.4 \mathrm{amp}
  2. B254×1019J,6.2amp\frac{25}{4} \times 10^{19} \mathrm{ J}, 6.2 \mathrm{amp}
  3. C252×1019J,0.8amp\frac{25}{2} \times 10^{19} \mathrm{ J}, 0.8 \mathrm{amp}
  4. DNone of these

Correct answer

A. 253×1019J,0.4amp\frac{25}{3} \times 10^{19} \mathrm{ J}, 0.4 \mathrm{amp}

Step-by-step solution

Pin =25W,λ=6600A˚=6600×10−10mnhv=PP_{\text{in }}=25 \mathrm{ W}, \lambda=6600 \AA=6600 \times 10^{-10} \mathrm{ m}n h v=P Number of photons emitted/sec n=Phcλ=Pλhc=25×6600×10−106.64×10−34×3×108=8.28×1019=253×1019\begin{aligned} n &=\frac{P}{\frac{h c}{\lambda}}=\frac{P \lambda}{h c}=\frac{25 \times 6600 \times 10^{-10}}{6.64 \times 10^{-34} \times 3 \times 10^{8}} \\ &=8.28 \times 10^{19}=\frac{25}{3} \times 10^{19} \end{aligned}3 \%ofemittedphotonsareproducingcurrentof emitted photons are producing current∴I=3100×ne=3100×253×1019×1.6×10−19=0.4A\begin{aligned} \therefore \quad & I=\frac{3}{100} \times n e \\ &=\frac{3}{100} \times \frac{25}{3} \times 10^{19} \times 1.6 \times 10^{-19}=0.4 \mathrm{ A} \end{aligned}$
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