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 the emitted electrons from the surface of the material to infinite distance. The frequency of some light is supposed to be \(\nu\). 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:

\[  h\nu = \phi + T  \]

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.

\[  \phi = h\nu = \frac{hc}{\lambda}  \]

\[       = \frac{6.63\times 10^{-34}\cdot 3.00\times 10^8}{542\times 10^{-9}}  \]

\[      = 3.67 \times 10^{-19} \ \mathrm{J}  \]

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 \(h\nu = \phi + T\), we have the kinetic energy.

\[  T = h\nu’ – \phi = 2.76 – 2.29 = 0.47 \ \mathrm{eV}  \]