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A 250 W electric bulb of 80% efficiency ...

A 250 W electric bulb of 80% efficiency emits a light of 6626 À wavelength. The number of photons emitted per second by the lamp is `(h = 6.636 xx 10 ^(-34) Js)`

A

`1.42 xx 10 ^(17)`

B

`2.18 xx 10 ^(16)`

C

`6.66 xx 10 ^(20)`

D

`2.83 xx 10 ^(16)`

Text Solution

Verified by Experts

The correct Answer is:
C
Energy (power) of the electric bulb = 250 W But it is only 80% efficient,
thus its actual energy `= ( 250 xx 80)/(100) = 200 W`
Energy of one photon, `E =( bc)/(lamda)`
`= (6.626 xx 10 ^(-34) xx 3 xx 10 ^(8))/( 6626 xx 10 ^(-10)) = 3 xx 10 ^(-19)`
Number of photons emited per second
`= ("Energy of the source ")/("Energy of one photon")= ( 200)/(3 xx 10 ^(-19)) = 66.66 xx 10 ^(19)`
`= 6.66 xx 10 ^(20)`
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