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Physics Question on de broglie hypothesis

The number of photons emitted per second by a 60 W60\, W source of monochromatic light of wavelength 663 nm663\, nm is (h=6.63×10−34Js)(h=6.63\times {{10}^{-34}}Js)

A

4×10−204\times {{10}^{-20}}

B

1.5×10201.5\times {{10}^{20}}

C

3×10−203\times {{10}^{-20}}

D

2×10202\times {{10}^{20}}

Answer

3×10−203\times {{10}^{-20}}

Explanation

Solution

Energy, E=nhcλE=\frac{nhc}{\lambda }
⇒\Rightarrow 60×1 Js=n×6.63×10−34Js×3×108m663×10−9m60\times 1\,Js=\frac{n\times 6.63\times {{10}^{-34}}Js\times 3\times {{10}^{8}}m}{663\times {{10}^{-9}}m}
[∵power=energytime]\left[ \because power=\frac{energy}{time} \right]
∴\therefore n=60×1×663×10−96.63×10−34×3×108n=\frac{60\times 1\times 663\times {{10}^{-9}}}{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}
=2×1020=2\times {{10}^{20}}