## College Physics (4th Edition)

The maximum wavelength of photons that will eject electrons from this metal is $507~nm$
We can find the magnitude of the energy removed from the electron by the stopping potential: $E = q~\Delta V$ $E = (1.6\times 10^{-19}~C)(1.10~V)$ $E = 1.76\times 10^{-19}~J$ This energy is equal to the maximum kinetic energy of the electron as it is ejected from the metal plate. We can find the work function $\phi$: $K_{max} = \frac{hc}{\lambda}-\phi$ $\phi = \frac{hc}{\lambda}-K_{max}$ $\phi = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~/s)}{350\times 10^{-9}~m}-1.76\times 10^{-19}~J$ $\phi = 5.679\times 10^{-19}~J-1.76\times 10^{-19}~J$ $\phi = 3.92\times 10^{-19}~J$ We can find the threshold wavelength for this work function: $\phi = \frac{hc}{\lambda}$ $\lambda = \frac{hc}{\phi}$ $\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{3.92\times 10^{-19}~J}$ $\lambda = 5.07\times 10^{-7}~m$ $\lambda = 507~nm$ The maximum wavelength of photons that will eject electrons from this metal is $507~nm$