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Question: The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100W is (...

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

A

100

B

1000

C

3×10203 \times 10 ^ { 20 }

D

3×10183 \times 10 ^ { 18 }

Answer

3×10203 \times 10 ^ { 20 }

Explanation

Solution

By using n=Pλhc=100×540×10−96.6×10−34×3×108=3×1020n = \frac { P \lambda } { h c } = \frac { 100 \times 540 \times 10 ^ { - 9 } } { 6.6 \times 10 ^ { - 34 } \times 3 \times 10 ^ { 8 } } = 3 \times 10 ^ { 20 }