Answer
(C) $5.92W$
Work Step by Step
Power produced by $\lambda_1$ is given as
$P_1=\frac{(1.49\times 10^{19}photons/s)}{2}\frac{hc}{\lambda_1}$
We plug in the known values to obtain:
$P_1=\frac{(1.49\times 10^{19})(6.63\times 10^{-34})(3\times 10^8)}{(2)488\times 10^{-9}}$
$P_1=3.035W$
Now for $\lambda_2$
$P_2=\frac{(1.49\times 10^{19})(6.63\times 10^{-34})(3\times 10^8)}{(2)514.5\times 10^{-9}}=2.88W$
Total power $=P_1+P_2$
Total power $=3.035W+2.88W=5.92W$
Thus, option (C) is correct.
