Question: Violet light of wavelength \[{\rm{400 nm}}\] ejects electrons with a maximum kinetic energy of \[{\rm{0}}{\rm{.860 eV}}\] from sodium metal. What is the binding energy of electrons to sodium metal?

Given,

Wavelength is, \(\lambda = 400\,{\rm{nm}} = 400 \times {10^{ - 9}}{\rm{m}}\).

Kinetic energy is, \({\rm{KE}} = 0.860\,{\rm{eV}}\).

We also know that: Planks constant \(h = 4.13 \times {10^{ - 15}}\,{\rm{eV}}{\rm{.s}}\)

Speed of light \(c = 3 \times {10^8}\,{\rm{m/s}}\)

The kinetic energy of the electron is given by

\(K{E_e} = hf - BE\) ...(1)

Here\(K{E_e}\)is the kinetic energy,\(h\)is the plank constant,\(f\) is the frequency of the EM radiation and\(BE\)is the binding energy.

Now we know that the wavelength of EM radiation is given by

\(\lambda = \frac{c}{f}\) ...(2)

Where\(c\)is the speed of light.

So equation becomes,

\(K{E_e} = \frac{{hc}}{\lambda } - BE\) ...(3)

Hence the binding energy is expressed as,

\(BE = \frac{{hc}}{\lambda } - KE\)

Substitute all the value in the above equation

\(\begin{align}{c}BE = \dfrac{{\left( {4.13 \times {{10}^{ - 15}}\,{\rm{eV}}{\rm{.s}}} \right)\left( {3.00 \times {{10}^8}\,{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}} \right)}}{{400 \times {{10}^{ - 9}}\,{\rm{m}}}} - (0.860\,{\rm{eV}})\\ &= 2.24\,{\rm{eV}}\end{align}\)

Therefore, the binding energy of electrons to sodium metal \(2.24\,{\rm{eV}}\).