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Q 11PE

Expert-verifiedFound in: Page 1063

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Question: 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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