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Expert-verifiedViolet 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?
The binding energy of electrons to sodium metal \(2.24\,{\rm{eV}}\)
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
\(\begin{aligned}{}BE &= \frac{{\left( {4.13 \times {{10}^{ - 15}}\,{\rm{eV}}{\rm{.s}}} \right)\left( {3.00 \times {{10}^8}\,{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}} \right)}}{{400 \times {{10}^{ - 9}}\,{\rm{m}}}} - (0.860\,{\rm{eV}})\\ &= 2.24\,{\rm{eV}}\end{aligned}\)
Therefore, the binding energy of electrons to sodium metal \(2.24\,{\rm{eV}}\)
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