Question

Calculate the binding energy in eV of electrons in aluminium, if the longest-wavelength photon that can eject them is \({\rm{304 nm}}\).

Short answer

Binding energy in eV of electrons in aluminium is, \(BE = 4.08\,{\rm{eV}}.\)

Step-by-step solution

The longest-wavelength EM radiation can eject a photoelectron

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)

Determination of the binding energy

For wavelength to be maximum, the kinetic energy of the electron must be zero. Hence we have

\(\begin{aligned}{}0 &= \frac{{hc}}{\lambda } - BE\\BE &= \frac{{hc}}{\lambda }\end{aligned}\)

Here, the longest-wavelength of the incident photon that can eject electrons from aluminium is,

\(\begin{aligned}{}\lambda &= 304\,{\rm{nm}}\\ &= 304 \times {10^{ - 9}}\,{\rm{m}}.\end{aligned}\)

We also know:

\(\begin{aligned}{}h &= 6.63 \times {10^{ - 34}}\,{\rm{J}}{\rm{.s}}\\c &= 3 \times {10^8}\,{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\end{aligned}\)

Substitute all the values in the above equation (1), and we get:

\(\begin{aligned}{}BE &= \frac{{\left( {6.62 \times {{10}^{ - 34}}\,{\rm{J \times s}}} \right)\left( {3 \times {{10}^8}\,{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right)}}{{304 \times {{10}^{ - 9}}\,{\rm{m}}}}\\ &= 6.53 \times {10^{ - 19}}\,{\rm{J}} \times \frac{{1\,{\rm{eV}}}}{{1.60 \times {{10}^{ - 19}}\,{\rm{J}}}}\\ &= 4.08\,{\rm{eV}}\end{aligned}\)

Hence, the binding energy in eV of electrons in aluminium is\(BE = 4.08\,{\rm{eV}}.\)