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Q 12PE

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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: What 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.

See the step by step solution

## Step 1: 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 2: Wavelength of the radiation

The total energy of the photon is

$$\begin{array}{c}E = (2.00\,{\rm{eV}}) + (2.71\,{\rm{eV}})\\ = 4.71\,{\rm{eV}}\end{array}$$

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

$$\begin{array}{c}\lambda = \frac{{\left( {4.13 \times {{10}^{ - 15}}\,{\rm{eV}}{\rm{.s}}} \right)\left( {3 \times {{10}^8}\,{\rm{m/s}}} \right)}}{{(4.71\,{\rm{eV}})}}\\ = 263 \times {10^{ - 9}}\,{\rm{m}}\end{array}$$

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.