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Q10PE

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College Physics (Urone)
Found in: Page 1063

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Short Answer

Violet 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 \(2.24\,{\rm{eV}}\)

See the step by step solution

Step by Step Solution

Step 1: Given data

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 \(c = 3 \times {10^8}\,{\rm{m/s}}\)

Step 2: 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)

Step 3: Calculate the binding energy of electrons to sodium metal

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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