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Q 11PE

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Found in: Page 1063

### College Physics (Urone)

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.

See the step by step solution

## Step 1: Given data

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

## 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 uranium metal

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 $$3.63\,{\rm{eV}}$$.