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CBSE Class 12 Physics 2023 Outside Delhi Set 3 Paper

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Question : 1 of 13
Marks: +1, -0
SECTION - A
The energy of a photon of wavelength 663 nm663\ \mathrm{nm} is
Solution:     👈: Video Solution
E  =  hcλE\;=\;\frac{hc}{\lambda}
  =  6.6×10−34×3×108663×10−9\;=\;\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{663 \times 10^{-9}}
  =2.98×10−19 J\;=2.98 \times 10^{-19}\ \mathrm{J}
    ≈3.0×10−19 J\;\; \approx 3.0 \times 10^{-19}\ \mathrm{J}
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