## chapter 1 , question 23

$\lambda=\frac{h}{p}$

004744172
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Joined: Fri Jun 23, 2017 11:39 am

### chapter 1 , question 23

with the energy given as 140.511 keV, what are the series of formulas that need to be used in order to calculate the wavelength?
i originally tried E=h(frequency), then would convert frequency= speed of light/lambda, but units arent canceling out...

Daniela_Chem14A
Posts: 27
Joined: Tue Nov 15, 2016 3:00 am

### Re: chapter 1 , question 23

E=hv ; v=frequency
c=(lambda)v

lambda= hc/E = [(Js)(m/s)]/J = m

Plank's constant with (Joules)(seconds) , Js, cancels out with Energy that equals Joules (J).
The seconds, s, cancels out with the s^-1 of speed of light.
At the end you are left with meters, m, which, are the units you need to get for lambda.

Hope this helps!

Diego Zavala 2I
Posts: 65
Joined: Fri Sep 29, 2017 7:07 am
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### Re: chapter 1 , question 23

You will also need the conversion factor of 1.602x10^-19J/eV (don't forget to convert kJ to joules)