Question

What is the longest-wavelength EM radiation that can eject a photoelectron from silver, given that the binding energy is 4.73 eV? Is this in the visible range?

Short answer

The longest wavelength of EM radiation is 263nm and the EM radiation is not in visible range.

Step-by-step solution

The longest-wavelength EM radiation can eject a photoelectron

The kinetic energy of the electron is given by

KE_e = hf -BE ...(1)

HereKE_e is the kinetic energy, his the plank constant, f is the frequency of the EM radiation and is the binding energy.

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

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

Where cis the speed of light.

So equation becomes,

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

Determination of the maximum wavelength

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

The given value is

\begin{aligned}BE=4.73\,{\rm{eV}}\\=(4.73\,{\rm{eV}})\left({1.60\times{{10}^{19}}\,\frac{{\rm{J}}}{{{\rm{eV}}}}} \right)\\= 7.57 \times {10^{ - 19}}\,{\rm{J}}\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}

h = 6.63×10^-34 J s

c = 3×10⁸m/s

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

\begin{aligned}\lambda=\frac{{\left({6.63\times{{10}^{ -34}}\,{\rm{J}}{\rm{.s}}} \right)\left({3\times {{10}^8}\,{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right\{\rm{s}}}}\right)}}{{\left({7.57\times{{10}^{-19}}\,{\rm{J}}}\right)}}\\=2.63\times{10^{-7}}\,{\rm{m}}\\ = 263\,{\rm{nm}}\end{aligned}

Hence the maximum wavelength of the EM radiation that can remove electrons from silver is 263nm

Therefore the longest-wavelength EM radiation can eject a photoelectron 263

Hence, the EM radiation is not in the visible range.