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Q14PE

Expert-verifiedFound in: Page 1063

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** \(v = 1.95 \times {10^6}\,{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}\)**

**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

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

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

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