1.

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

Answer»

<P>`3XX10^(20)`
`1.5xx10^(20)`
`2xx10^(-20)`
`2xx10^(20)`

Solution :`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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