## Problem 1.33 part c

Shirley Wong 2E
Posts: 22
Joined: Fri Sep 25, 2015 3:00 am

### Problem 1.33 part c

I was able to do part a and b of 1.33 in the textbook, but I'm completely stumped on part c) What is the wavelength of the radiation that caused photoejection of the electron? Thank you!

Umair Khan 2G
Posts: 32
Joined: Fri Sep 25, 2015 3:00 am

### Re: Problem 1.33 part c

For part c, we use the formula for the photoelectric effect: E(photon) - $\Phi$ = E(kinetic)

Since part A gave us the velocity of the ejected electron we can find the E(kinetic) with E(k) =$\frac{1}{2}m^{2} v^{2}$, with m being the mass of the electron and v being the given velocity.
In part B we solved for the threshold (the energy required to remove the electron)
We can rearrange the formula to give us:E(photon) = E(kinetic) + $\Phi$

Solving this will give us the energy of the photon: E = h x v which is the same as $E = h\frac{c}{\lambda }$

From here we can solve for $\lambda$ to give us the wavelength.