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No .of photon emitted eery second from 6...

No .of photon emitted eery second from 60 W bulb will be ………

A

`3xx10^(20)`

B

`1.5xx10^(20)`

C

`2xx10^(-20)`

D

`2xx10^(20)`

Text Solution

Verified by Experts

The correct Answer is:
D
`P=(E )/(t)=(nhf)/(t)=(nhc)/(lambdat)`
`therefore n(plambdat)/(hc)=(60xx60xx10^(-8)xx1)/(6.6xx10^(-34)xx3xx10^(8))=2xx10^(20)`
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