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Q14PE

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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

What is the maximum velocity of electrons ejected from a material by $$80\,{\rm{nm}}$$ photons, if they are bound to the material by $$4.73\,{\rm{eV}}$$?

The maximum velocity of electrons $$v = 1.95 \times {10^6}\,{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}$$

See the step by step solution

Step 1: The maximum velocity of electrons

From equation, the maximum kinetic energy of the ejected electron from any metal surface is

$$K{E_e} = hf - BE$$ ...(1)

Where $$h = 6.626 \times {10^{ - 34}}\,{\rm{Js}}$$ is Planck's constant,

$$f$$is the frequency of the incident photon,

$$hf$$is the photon's energy, and

The relation between frequency $$f$$and wavelength $$\lambda$$is given by

$$f = \frac{c}{\lambda }$$ ...(2)

Therefore, from the above equation, we get

$$K{E_e} = \frac{{hc}}{\lambda } - BE$$ ...(3)

Now as we know the kinetic energy of an object having mass $$m$$ and velocity v is given by

$$K{E_e} = \frac{1}{2}m{v^2}$$ ...(4)

Here, binding energy of the electron in material is

$$BE = 4.73\,{\rm{eV}}$$

Since, we know

$$1.00\;\,{\rm{J}} = 6.242 \times {10^{18}}\,{\rm{eV}}$$

Therefore we get

\begin{aligned}{}BE &= 4.73\,{\rm{eV}} \times \frac{{1.00\,{\rm{J}}}}{{6.242 \times {{10}^{18}}\,{\rm{eV}}}}\\ &= 7.57 \times {10^{ - 19}}\,{\rm{J}}\end{aligned}

Step 2: Calculate the maximum velocity of electrons

Wavelength of the ejected electron is

\begin{aligned}{}\lambda &= 80\,{\rm{nm}}\\ &= 80 \times {10^{ - 9}}\,{\rm{m}}\end{aligned}

Mass of the electron is

$$m = 9.11 \times {10^{ - 31}}\,{\rm{kg}}$$

Now from equation, the kinetic energy of the ejected electron is

$$K{E_e} = \frac{{hc}}{\lambda } - BE$$

Substitute all the value in the above equation

\begin{aligned}{}K{E_e} &= \frac{{6.626 \times {{10}^{ - 34}}\,{\rm{Js}} \times 3.00 \times {{10}^8}\,{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}}}{{80 \times {{10}^{ - 9}}\,{\rm{m}}}} - 7.57 \times {10^{ - 19}}\,{\rm{J}}\\ &= 1.73 \times {10^{ - 18}}\,{\rm{J}}\end{aligned}

Hence, from equation, the velocity of the ejected electron is

$$v = \sqrt {\frac{{2K{E_e}}}{m}}$$

Substitute all the value in the above equation

\begin{aligned}{}v &= \sqrt {\frac{{2 \times 1.73 \times {{10}^{ - 18}}\,{\rm{J}}}}{{9.11 \times {{10}^{ - 31}}\,{\rm{kg}}}}} \\ &= 1.95 \times {10^6}\,{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}\end{aligned}

Hence, $$v = 1.95 \times {10^6}\,{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}$$