Light of wavelength 400 nm is incident upon lithium (\phi = 2.93 eV). Calculate (a) the photon energy and (b) the stopping potential V_{0}.
Strategy (a) Light is normally described by wavelengths in nm, so it is useful to have an equation to calculate the energy in terms of λ.
\begin{aligned}E &=h f=\frac{h c}{\lambda} \\&=\frac{\left(6.626 \times 10^{-34} J \cdot s \right)\left(2.998 \times 10^{8} m / s \right)}{\lambda\left(1.602 \times 10^{-19} J / eV \right)\left(10^{-9} m / nm \right)}\end{aligned}E=\frac{1.240 \times 10^{3} eV \cdot nm }{\lambda} (3.35)
(b) We use Equation (3.32) to determine the stopping potential once we know the frequency f and work function \phi.
\frac{1}{2} m v_{\max }^{2}=e V_{0}=h f-\phi (3.32)
