Answer
(a) red
(b) blue
(c)
red: $4.9\times 10^{20} photons/s$
blue: $5.8\times 10^{19} photons/s$
Work Step by Step
(a) We know that
$\frac{(\frac{P}{E})_{red}}{(\frac{P}{E})_{blue}}=\frac{\lambda_{red}P_{red}}{\lambda_{blue}P_{blue}}$
We plug in the known values to obtain:
$\frac{(\frac{P}{E})_{red}}{(\frac{P}{E})_{blue}}=\frac{(350\times 10^{-9}m)(150W)}{(460\times 10^{-9}m)(25W)}=8.5\gt 1$
As $(\frac{P}{E})_{red}\gt (\frac{P}{E})_{blue}$
Thus, the red bulb emits more photons per second as compared to the blue bulb.
(b) We know that
$\frac{E_{red}}{E_{blue}}=\frac{\lambda_{blue}}{\lambda_{red}}$
We plug in the known values to obtain:
$\frac{E_{red}}{E_{blue}}=\frac{460\times 10^{-9}m}{650\times 10^{-9}m}=0.71\lt 1$
Thus, $E_{red}\lt E_{blue}$ and we conclude that the blue bulb emits photons of higher energy than the red bulb.
(c) We know that the number of photons emitted per second from the red bulb is given as
$n=\frac{P\lambda}{hc}$
We plug in the known values to obtain:
$n=\frac{(150W)(650\times 10^{-9}m)}{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}=4.9\times 10^{20}photons/s$
Now the number of photons emitted per second from the blue bulb is given as
$n=\frac{P\lambda}{hc}$
We plug in the known values to obtain:
$n=\frac{(25W)(460\times 10^{-9}m)}{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}$
$\implies n=5.8\times 10^{19}photons/s$
