## Week 3 discussion, question 6 [ENDORSED]

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

Husnia Safi - 1K
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### Week 3 discussion, question 6

I need help answering this problem:
The work function for chromium metal is 4.37 eV. What wavelength of radiation must be used to eject electrons with a velocity of 1.5x10^3 km/s?

Chem_Mod
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### Re: Week 3 discussion, question 6

This is a photoelectric problem so use the photoelectric equation we have learned. Also the work function of the metal is 4.37eV. You need to convert to J. 1eV = 1.60 * 10^-19 J.

Rebecca Chu 1C
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Joined: Fri Apr 06, 2018 11:02 am

### Re: Week 3 discussion, question 6  [ENDORSED]

First, you have convert 4.37 eV to J so you have 4.37 eV x [(1.602 x 10^-19 J)/1 eV] = 7.0 x 10^-19 J. It also tells you that this is your work function so this is your $\Phi$
You should also convert v = 1.5 x 10^3 km/s to 1.5 x 10^6 m/s since we will be working with Joules.
Then you use $KE = \frac{1}{2} m_{-e}v_{-e}^{2}$ to find the kinetic energy so KE = (1/2)(9.1095 x 10^-31 kg)[(1.5 x 10^6 m/s)^2]. You end up with KE = 1.0248 x 10^-18 J. The units end up in Joules because 1 J = 1 kg(m^2)(s^-2)
Now that you have the KE and the work function, you can find the energy of the photon by using $E_{p} = KE + \Phi$ so E = (1.0248 x 10^-18 J) + (7 x 10^-19 J) = 1.7248 x 10^-18 J
To find the wavelength, you can then combine the E = hv and c = $\lambda$ v to get the equation $\lambda = (hc)/E_{p}$
This gives you $\lambda$ = [(6.626 x 10^-34 Js)(3.0 x 10^8)]/(1.7248 x 10^-18 J) = 1.15 x 10^-7 m
Convert to nanometers and you get 115 nm.

(FYI, this is question 1.34 from the textbook)