Homework Answers
30)
Given:
lambda = 475 NM = 4.75*10^-7 m
1st calculate energy of 1 photon
use:
E = h*c/lambda
=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(4.75*10^-7 m)
= 4.185*10^-19 J
number of photon = total energy/energy of 1 photon
n = 2.0E-17/4.185*10^-19
= 47.79
Answer: 48
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