## Speed in a Vacuum

$c=\lambda v$

Sean1F
Posts: 19
Joined: Sat Sep 07, 2019 12:16 am

### Speed in a Vacuum

When is the speed in a vacuum used? Does anybody have any broad examples of when it is used?

Alice Chang 2H
Posts: 101
Joined: Fri Aug 30, 2019 12:18 am

### Re: Speed in a Vacuum

Speed in a vacuum is defined as c= 3.00*10^8 m/s, which is a constant while measuring other speeds/variables in the electromagnetic spectrum topic.
A formula given from the lecture is $c=\lambda \times v$, which is (the speed of light) = (wavelength) x (frequency). You can use the speed of light, a constant, to find the wavelength of light, for example.

Here's the example given in the lecture (paraphrased):
3.61*10^-19 J is required to remove 1 electron without kinetic energy, what is the longest wavelength of light that can do this?
$v=c\div \lambda$ and $E=hv$, so replace v in $E=hv$ to be $E= h(c\div \lambda)$, which is then $\lambda = hc\div E$ to find lambda (which is wavelength of light). Then, plug in numbers to get (6.626 x 10^34 Js)(3.00 x 10^8 m/s)/(3.61 x 10^-19 J) = 551 nm (wavelength of light to remove 1 electron without kinetic energy)

This is a mess to explain online, but hope this helps somehow!

Sean1F
Posts: 19
Joined: Sat Sep 07, 2019 12:16 am

### Re: Speed in a Vacuum

Alice Chang 4B wrote:Speed in a vacuum is defined as c= 3.00*10^8 m/s, which is a constant while measuring other speeds/variables in the electromagnetic spectrum topic.
A formula given from the lecture is $c=\lambda \times v$, which is (the speed of light) = (wavelength) x (frequency). You can use the speed of light, a constant, to find the wavelength of light, for example.

Here's the example given in the lecture (paraphrased):
3.61*10^-19 J is required to remove 1 electron without kinetic energy, what is the longest wavelength of light that can do this?
$v=c\div \lambda$ and $E=hv$, so replace v in $E=hv$ to be $E= h(c\div \lambda)$, which is then $\lambda = hc\div E$ to find lambda (which is wavelength of light). Then, plug in numbers to get (6.626 x 10^34 Js)(3.00 x 10^8 m/s)/(3.61 x 10^-19 J) = 551 nm (wavelength of light to remove 1 electron without kinetic energy)

This is a mess to explain online, but hope this helps somehow!

Thanks!
This helps a lot! :D