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### UA Session Question

Posted: Wed Oct 28, 2020 10:27 pm
Hi everyone! Could somebody help me with this question?

Q8. You have a metal, Rubidium, which has a work function of 2.3 eV. If the electrons being
ejected from the metal have a wavelength of 1.7 x 10^5 m, what is the wavelength of the incident
light?

### Re: UA Session Question

Posted: Wed Oct 28, 2020 10:33 pm
Since the wavelength of the electrons is given (1.7 x 10^5 m), you can first find the velocity of the ejected electrons by using De Broglie's wave equation: $\lambda = \frac{h}{mv}$ . After finding velocity of the ejected electron, you can find the kinetic energy of the ejected electron ( $\frac{1}{2}mv^{2}$ ). Then, you can find the wavelength of the incident light by using the equation $E_{photon} - \Phi (work function) = E_{k}$ , where $E_{photon} = \frac{hc}{\lambda }$ , and $E_{k}$ is the kinetic energy you calculated previously. In this equation, you are solving for $\lambda$, which is the wavelength of the incident light. I hope this helps!

### Re: UA Session Question

Posted: Fri Oct 30, 2020 4:09 pm
Hi Nina!
To solve this problem, you need to use the general equation for the photoelectric effect, which is: energy of photon = work function + kinetic energy of electron
1. Convert work function of 2.3 eV to joules: 1.436 x 10^19 J
2. Use the De Broglie equation to solve for the velocity of the electron: 1.7 x 10^5 = (6.626 x 10^-34)/(9.11 x 10^-31 x v) solve for v
3. Use the kinetic energy equation (Ke = 0.5mv^2) to find the energy of the ejected electron (use v solved in above equation)
4. Add the work function and kinetic energy of the electron to find the energy of the photon
4. Use E=(hc)/lambda to find the wavelength of the incident light