## Worked Example

604807557
Posts: 51
Joined: Wed Nov 16, 2016 3:02 am

### Worked Example

I'm going through the "The Quantum World" notes in the course reader and after reviewing the first worked example:
"The following questions relate to the same metal used in a series of photoelectric experiments.
If 3.61 x 10^-19 J is required to remove an electron with zero kinetic energy from a metal surface, what would be the longest wavelength light that could do this?" I don't remember how combining E = hv and c = λv gives E = hc/λ. How do you reach that final formula (E = hc/λ) in order to solve for the wavelength?

Sarah_Wilen
Posts: 62
Joined: Fri Jun 23, 2017 11:39 am
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### Re: Worked Example

Here are your two equations to combine:
$E=h\nu$
$c=\lambda \nu$

You set the speed of light equation to equal velocity:
$c=\lambda \nu \rightarrow \frac{c}{\lambda }=\frac{\lambda \nu }{\lambda }\rightarrow \frac{c}{\lambda }=\nu$

Then, you see that $E=h\nu$ has velocity in it, so you can plug the velocity from the previously solved equation^ into the velocity in the energy equation.
$E=\frac{hc}{\lambda }$

Yay! We have that equation, now we can conquer the world or solve the rest of the question.

Goal: You want to find the wavelength using the equation: $E=\frac{hc}{\lambda }$ and we are given the energy of the incident light (3.61 x 10^-19 J).

1) You can rearrange the equation to soluve for wavelength:
$E=\frac{hc}{\lambda }\rightarrow E\left ( \lambda \right )=\frac{hc}{\lambda }\left ( \lambda \right )\rightarrow \frac{E\lambda }{E}=\frac{hc}{E}\rightarrow \lambda =\frac{hc}{E}$
2) Now, you have the energy, planck's constant, and speed of light constant to solve for wavelength. Plug and chug:
$\lambda =\frac{hc}{3.61\cdot 10^{-19} J}$
3)**Sometimes you would multiply the wavelength by 10^9 m to convert meters to nanometers for convenience.

Rejoice!