Question

UV radiation having a $300-\mathrm{n}\mathrm{m}$ wavelength falls on uranium metal, ejecting $0.500\mathrm{e}\mathrm{V}$ electrons. What is the binding energy of electrons to uranium metal?

Short answer

The binding energy of electrons to uranium metal $3.63\mathrm{e}\mathrm{V}$

Step-by-step solution

Given data

Given,

Wavelength is, $\lambda =300\mathrm{n}\mathrm{m}=300\times {10}^{-9}\mathrm{m}$.

Kinetic energy is, $\mathrm{K}\mathrm{E}=0.500\mathrm{e}\mathrm{V}$.

We also know that: Planks constant $h=4.13\times {10}^{-15}\mathrm{e}\mathrm{V}.\mathrm{s}$

Speed of light $c=3\times {10}^8\mathrm{m}/\mathrm{s}$

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)

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

$\begin{array}{rl}BE& =\frac{\left(4.13\times {10}^{-15}\mathrm{e}\mathrm{V}.\mathrm{s}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{\left(300\times {10}^{-9}\mathrm{m}\right)}-(0.500\mathrm{e}\mathrm{V})\\ & =3.63\mathrm{e}\mathrm{V}\end{array}$

Therefore, the binding energy of electrons to uranium metal $3.63\mathrm{e}\mathrm{V}$