(a) What frequency of light is needed to produce electrons of kinetic energy 3.00 eV from illumination of lithium? (b) Find the wavelength of this light and discuss where it is in the electromagnetic spectrum.
Strategy We have enough information to determine the photon energy needed from Equation (3.30), and we can determine the frequency from E = hf.
h f=\phi+\frac{1}{2} m v_{\max }^{2} (3.30)
From Equation (3.30), we have
\begin{aligned}h f &=\phi+\frac{1}{2} m v_{\max }^{2} \\&=2.93 eV +3.00 eV =5.93 eV\end{aligned}
The photon frequency is now found to be
\begin{aligned}f=\frac{E}{h} &=\frac{(5.93 eV )\left(1.60 \times 10^{-19} J / eV \right)}{\left(6.626 \times 10^{-34} J \cdot s \right)} \\&=1.43 \times 10^{15} s ^{-1}=1.43 \times 10^{15} Hz\end{aligned}
(b) The wavelength of the light can be found from c=\lambda f.
\lambda=\frac{c}{f}=\frac{3.00 \times 10^{8} m / s }{1.43 \times 10^{15} Hz }=2.10 \times 10^{-7} m =210 nm
This is ultraviolet light, because the wavelength 210 nm is below the range of visible wavelengths 400 to 700 nm.