Question

Question: UV radiation having a 300nmwavelength falls on uranium metal, ejecting 0.500eVelectrons. What is the binding energy of electrons to uranium metal?

Short answer

The binding energy of electrons to uranium metal 3.36 eV.

Step-by-step solution

Given data

Given,

Wavelength is, \(\lambda = 300\,{\rm{nm}} = 300 \times {10^{ - 9}}{\rm{m}}\).

Kinetic energy is, \({\rm{KE}} = 0.500\,{\rm{eV}}\).

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 longest-wavelength EM radiation can eject an electron

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)

Calculate the binding energy of electrons to uranium metal

The binding energy is expressed as,

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

Substitute all the value in the above equation

\(\begin{aligned}{c}BE = \frac{{\left( {4.13 \times {{10}^{ - 15}}\,{\rm{eV}}{\rm{.s}}} \right)\left( {3 \times {{10}^8}\,{\rm{m/s}}} \right)}}{{\left( {300 \times {{10}^{ - 9}}\,{\rm{m}}} \right)}} - (0.500\,{\rm{eV}})\\ &= 3.63\,{\rm{eV}}\end{aligned}\)

Therefore, the binding energy of electrons to uranium metal \(3.63\,{\rm{eV}}\).