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Q10PE

Expert-verifiedFound in: Page 1063

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

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

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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