CBSE Class 12-science Answered
an electric bulb of power W is emitting light of a particular color of wavelength lamda. the number of photons emitted per second is-?
Asked by Abhijeet | 29 Apr, 2013, 06:23: PM
Expert Answer
P=W
=energy/time
=N (hc/lambda)/time......................hc/lambda=energy associated with one photon, N=number of photons emitted per second.
W=N (hc/lambda)/time.....time=1s
N=(W x lambda)/(hc)
Answered by | 30 Apr, 2013, 02:49: PM
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