How Do Red and Blue Laser Photon Emissions Compare at Equal Power?

[elephantorz]

[SOLVED] I'm doing something wrong...

1. A red laser with a wavelength of 650 nm and a blue laser with a wavelength of 450 nn emit laser beams with the same light power. How do their rates of photon emission compare? Answer this by computing R_{red} / R_{blue}
2. P = Rhf = dNhf / dt = dE_{light} / dt
R = dN / dt
E_{light} = Nhf, where N is the number of photons
f = c / \lambda
h = 6.624E-32 Js
3. I have the f_{red} = 4.62 x 10^{14} s
and f_{blue} = 6.67 x 10^{14} s , it says their Powers are the same, so I go ahead and went and equaled:

P_{red} to P_{blue}, which is:P_{red}h_{red}f_{red} = P_{blue}h_{blue}f_{blue}

but I am tired and can't see the relevance, because when I multiplied times h [in Js] and then I divide like explained, I get .6925, and the ANSWER is 1.44.

Help? Please?

 

[berkeman]

Since red photons carry less energy per photon, there will need to be more of them. You just got your answer upside-down.

Just use your equation for E = Nhf, and be sure to take the ratio in the correct direction.

 

[Chi Meson]

1. A red laser with a wavelength of 650 nm and a blue laser with a wavelength of 450 nn emit laser beams with the same light power. How do their rates of photon emission compare? Answer this by computing R_{red} / R_{blue}



2. P = Rhf = dNhf / dt = dE_{light} / dt
R = dN / dt
E_{light} = Nhf, where N is the number of photons
f = c / \lambda
h = 6.624E-32 Js



3. I have the f_{red} = 4.62 x 10^{14} s
and f_{blue} = 6.67 x 10^{14} s , it says their Powers are the same, so I go ahead and went and equaled:

P_{red} to P_{blue}, which is:


P_{red}h_{red}f_{red} = P_{blue}h_{blue}f_{blue}

but I am tired and can't see the relevance, because when I multiplied times h [in Js] and then I divide like explained, I get .6925, and the ANSWER is 1.44.

Help? Please?

Yes the answer is 1.44.

You got confused because you plugged in your number way too soon. Consider this:

Only the wavelengths are important. Obviously "R" here is used as "N," th number of photons per second? Well, P=P, you got that, and P=(Rhf)/t and f =c/ lamda.

Just substitute, see what cancel out, and use what you are left with.

 

[Kaleb]

I double checked the math, looks good.

 

[elephantorz]

Ok...that was...really weird, I guess I somehow reversed wavelengths? I simply did the reciprocal and it worked, thanks all!