## 1B 15C

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### 1B 15C

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 x 10^3 km/s.
(c) What is the wavelength of the radiation that caused photoejection of the electron?

I was able to find the wavelength of the ejected electron and the energy required to remove it from the metal. However, I don't know how to use it to solve for the wavelength of the radiation. The answer in the book says that you use 1/2mv^2 + the energy required to remove the electron from the metal to find the energy of the photon, and I don't understand why.

JChen_2I
Posts: 107
Joined: Fri Aug 09, 2019 12:17 am

### Re: 1B 15C

We know that the E(photon)=threshold energy + E(kinetic)
Threshold energy is the amount of energy required to remove an electron from a metal which is what we found in (b) to be 1.66*10^-17 J
E(kinetic)=1/2mv^2 where m=9.11*10^-31 kg (mass of electron) and v=3.6 * 10^3 km/s or 3.6*10^6m/s
so E(kinetic)=1/2(9.11*10^-31 kg)(3.6*10^6m/s) = 5.903*10^-18 J
then to find E(photon) you do 1.66*10^-17 J + 5.903*10^-18 J = 2.2503*-17 J
To find the wavelength, you use lambda=hc/E(photon)
so lambda=(6.626*10-34m^2kg/s)(3.0*10^8m/s)/2.2503*-17 J = 8.8 * 10^-9 m or 8.8 nm