Work function

Question:
450-nm (450$\times$10$^{-9}$ m) wavelength light is incident on sodium surface for which the threshold wavelength of the photoelectrons is 542 nm (542$\times$10$^{-9}$ m). Find the work function of sodium and kinetic energy of the incident light after photoelectrons released.

Answer:
The work function is defined as the minimum energy to obtain electrons from the material surface to infinite distance. The frequency of some light is $\nu$. The partial energy of light may be used for the minimum threshold, which is the work function, $\phi$. The extra energy becomes the kinetic energy, $T$. Namely, we have the following relationship:
\begin{equation}
h\nu = \phi + T
\end{equation}
The Planck's constant, $h$, is 6.63 $\times$ 10$^{-34}$ Js. The frequency, $\nu$, is given by the speed of light, $c$, divided by wavelength, $\lambda$. In this case, the threshold wavelength is given, so we can exclude the term of the kinetic energy.
\begin{eqnarray}
\phi &=& h\nu = \frac{hc}{\lambda}  \\
       &=& \frac{6.63\times 10^{-34}\cdot 3.00\times 10^8}{542\times 10^{-9}}  \\
      &=& 3.67 \times 10^{-19} \ \mathrm{J}
\end{eqnarray}
One joule is 6.24 $\times$ 10$^{18}$ electronvolts. Thus, we can convert it into eV.
\[
\phi = 3.67 \times 10^{-19} \times 6.24 \times 10^{18} = 2.29 \ \mathrm{eV}
\]
If we use 450-nm light, we have the total energy as
\[
h\nu' = \frac{6.63\times 10^{-34}\cdot 3.00\times 10^8}{450\times 10^{-9}}=4.42\times 10^{-19} \ \mathrm{J}
\]
Namely, it is 2.76 eV. Using equation (1), we have the kinetic energy.
\[
T = h\nu' - \phi = 2.76 - 2.29 = 0.47 \ \mathrm{eV}
\]

Powered by Hirophysics.com