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### Photoelectric Effect

Posted: Thu Oct 13, 2016 11:47 pm
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?

### Re: Photoelectric Effect

Posted: Fri Oct 14, 2016 9:51 am
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.

### Re: Photoelectric Effect

Posted: Fri Oct 14, 2016 10:29 am
But how does multiplying nm times nm cancel out nm to become meters??? wouldn't it have to be divided (m/nm) ?

### Re: Photoelectric Effect

Posted: Fri Oct 14, 2016 11:30 am
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.

### Re: Photoelectric Effect

Posted: Sat Jul 01, 2017 9:10 pm
Under which circumstances do we utilize the formula E=hc/lambda that incorporates the lambda?

### Re: Photoelectric Effect

Posted: Sun Jul 02, 2017 3:12 pm
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).