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Expert-verifiedWhat is the wavelength of EM radiation that ejects \(2.00\,{\rm{eV}}\) electrons from calcium metal, given that the binding energy is \(2.71\,{\rm{eV}}\)? What type of EM radiation is this?
Wavelength of the radiation is \(263 \times {10^{ - 9}}\,{\rm{m}}\)
Type of EM radiation is Ultra violet.
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 total energy of the photon is
\(\begin{aligned}{}E &= (2.00\,{\rm{eV}}) + (2.71\,{\rm{eV}})\\ &= 4.71\,{\rm{eV}}\end{aligned}\)
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 wavelength of the radiation is expressed as,
\(\lambda = \frac{{hc}}{E}\)
Substitute all the value in the above equation
\(\begin{aligned}{}\lambda &= \frac{{\left( {4.13 \times {{10}^{ - 15}}\,{\rm{eV}}{\rm{.s}}} \right)\left( {3 \times {{10}^8}\,{\rm{m/s}}} \right)}}{{(4.71\,{\rm{eV}})}}\\ &= 263 \times {10^{ - 9}}\,{\rm{m}}\end{aligned}\)
Therefore, the wavelength of the radiation is \(263 \times {10^{ - 9}}\,{\rm{m}}\)
Hence, the wavelength of the EM radiation is \(263\,{\rm{nm,}}\)the radiation type is ultra violet.
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