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Expert-verifiedQuestion: UV radiation having a 300nm wavelength falls on uranium metal, ejecting 0.500eV electrons. What is the binding energy of electrons to uranium metal?
The binding energy of electrons to uranium metal 3.36 eV.
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 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)
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 \(3.63\,{\rm{eV}}\).
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