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Q.

The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100 W is (taking h=6×10−34J−sec)

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a

100

b

1000

c

3×1020

d

3×1018

answer is C.

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Detailed Solution

p=nhcλt⇒100=n×6×10−34×3×108540×10−9×1⇒n=3×1020

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