2 Ultra-violet light of wavelength $\lambda$ incident on an emitter
3 surface gives rise to photoelectrons with maximum kinetic energy
4 $2.00\U{eV}$ whereas, for a wavelength $3\lambda/4$, the maximum
5 kinetic energy of emitted photoelectrons from the same surface is
6 $3.47\U{eV}$. \Part{a} Determine the wavelength $\lambda$ (in nm) of
7 these ultraviolet light waves incident on the emitter
8 surface. \Part{b} What are the photoelectric work function $\phi$ (in
9 eV) and the threshold wavelength $\lambda_\text{threshold}$ (in nm) of the
10 photons for photoelectron emission from this emitter surface. \Part{c} If
11 the wavelength of incident violet light is $400\U{nm}$, find the maximum
12 kinetic energy (in eV) of emitted photoelectrons.
17 Balancing energy in both cases
19 \phi &= \frac{hc}{\lambda} - K_\text{max} = \frac{4hc}{3\lambda} - K_\text{max}' \\
20 K_\text{max}' - K_\text{max} &= \frac{1}{\lambda}(\frac{4}{3}hc-hc) \\
21 \lambda &= \frac{hc}{3(K_\text{max}' - K_\text{max})}
23 \frac{3}{4}\lambda = \ans{211\U{nm}}
28 \phi &= \frac{hc}{\lambda} - K_\text{max}
30 \phi &= \frac{hc}{\lambda_\text{threshold}} \\
31 \lambda_\text{threshold} &= \frac{hc}{\phi}
37 K_\text{max} = \frac{hc}{\lambda} - \phi