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JEE Main 2019 (10 Apr, Shift 2) · Photons: Energy, Momentum & Radiation Pressure

Photons: Energy, Momentum & Radiation Pressure question from JEE Main 2019 (10 Apr, Shift 2)

Q30JEE Main 2019 · 10 APR · SHIFT 2Easy

Photons: Energy, Momentum & Radiation Pressure

A 2 mW laser operates at wavelength of 500 nm. The number of photons that will be emitted per second is: [Given Planck's constant h = 6.6 × 10–34 Js, speed of light c = 3.0 × 108 m/s]

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Answer: (a)

  1. Photons per second n=Phc/λ=Pλhcn = \dfrac{P}{hc/\lambda} = \dfrac{P\lambda}{hc}
  2. n=2×10−3×500×10−96.6×10−34×3×108=10−91.98×10−25n = \dfrac{2\times10^{-3}\times 500\times10^{-9}}{6.6\times10^{-34}\times 3\times10^{8}} = \dfrac{10^{-9}}{1.98\times10^{-25}}
  3. n≈5×1015n \approx 5\times10^{15} per second
  4. So the answer is (a) 5×10155\times10^{15}.
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