## Photoelectric Effect

JulissaLopez1F
Posts: 12
Joined: Wed Sep 21, 2016 2:57 pm

### Photoelectric Effect

When plugging into E=hc/lambda, and a incoming wavelength is given, do you take that wavelength and plug it into the denominator? Because in an example from my discussion section the given wavelength was plugged in but was also multiplied by (10^-9) and I have no idea why...
I know 1 m is 10^9 nm, does that have anything to do with it?

Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times

### Re: Photoelectric Effect

For me to be certain, I would need more information. But if I am right, the given wavelength is in nm, which you need to convert to meters, because the given c constant (speed of light (c) = m/s) is in SI unit for the calculation to work. So this should explain the 10-9 conversion factor.

JulissaLopez1F
Posts: 12
Joined: Wed Sep 21, 2016 2:57 pm

### Re: Photoelectric Effect

But how does multiplying nm times nm cancel out nm to become meters??? wouldn't it have to be divided (m/nm) ?

Anna_Kim_2E
Posts: 31
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Photoelectric Effect

I think it is nanometer that has been converted to meters.
When you want to convert, for example, 900nm into meters, because there are 1x10^9nm in 1 meter, or 1x10^-9m in 1nm , it would 900nm would be multiplied by 10^-9 to convert into meters.

derek1d
Posts: 19
Joined: Wed Jun 28, 2017 3:00 am

### Re: Photoelectric Effect

Under which circumstances do we utilize the formula E=hc/lambda that incorporates the lambda?

Joe Rich 1D
Posts: 32
Joined: Fri Jun 23, 2017 11:39 am
Been upvoted: 1 time

### Re: Photoelectric Effect

We utilize the formula E=hc/λ if we are given the wavelength of light and want to solve for the energy per photon of the light (or if we are given the energy per photon of the light and want to calculate the wavelength). This equation is really just combining the two equations c=λν and E=hν in order to save a step of p=using different equations (because if c=λν, then ν=c/λ).