## 1B.15 Part C

Rachel Yoo 1F
Posts: 65
Joined: Fri Sep 28, 2018 12:24 am

### 1B.15 Part C

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6*10^3 km*s^-1.
part c. what is the wavelength of the radiation that caused photoejection of the electron?
I got part a and b but I'm not really sure how to do this part but I tried both the equations lambda=c/v and lambda=ch/(work function) but I didn't get the answer on the back of the book

Maya_Peterson1C
Posts: 62
Joined: Fri Sep 28, 2018 12:28 am

### Re: 1B.15 Part C

I don't see a question 1B.15 in the textbook. Is it at the end of the chapter?

Jonathan Omens 1K
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

### Re: 1B.15 Part C

Use the equation to find the energy of the incoming radiation: $E_{photon}=E_{K}+\phi$. The velocity of the ejected electron is given and the mass of the electron is in the list of constants, so you can calculate $E_{K}$ (kinetic energy of electron) using $\frac{1}{2}mv^2$. In part (b) you calculated the work function ($\phi$). Plug these two values into the equation to get $E_{photon}$.
Now that you have the energy of the radiation (photon), calculate its wavelength ($\lambda$) using the functions $E=h\nu$ and $c=\nu \lambda$ (or just combine them to make $E=hc\lambda ^{-1}$).