# An electron is accelerated through an electric potential to a kinetic energy of 18.2 keV. What is...

## Question:

An electron is accelerated through an electric potential to a kinetic energy of 18.2 keV. What is its characteristic wavelength? (Hint: Recall that the kinetic energy of a moving object is E = 1/2 mv{eq}^2 {/eq} where m is the mass of the object and v is the speed of the object.)

## Kinetic Energy:

Kinetic energy possessed by an electron in classical mechanics can be depicted by the formula {eq}K.E = \dfrac{1}{2}m{v^2} {/eq}. This energy can also be used to estimate the value of electron's wavelength by using its mass.

Given Data:

• The kinetic energy of an electron is 18.2 keV.

The velocity of the electrons can be calculated as shown below.

{eq}K.E = \dfrac{1}{2}m{v^2} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( {\rm{I}} \right) {/eq}

Where;

• {eq}K.E {/eq} is the kinetic energy of electrons.
• {eq}m {/eq} is the mass of electrons.
• {eq}v {/eq} is the velocity of electrons.

The kinetic energy is in keV.

The conversion of keV into Joules is as shown below.

{eq}1\;{\rm{keV}} = {\rm{1}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{ - 16}}\;{\rm{J}} {/eq}

The conversion of 18.2 keV into Joules is done as shown below.

{eq}\begin{align*} 18.2\;{\rm{keV}} &= \left( {18.2 \times {\rm{1}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{ - 16}}} \right){\rm{J}}\\ &= {\rm{29}}{\rm{.12}} \times {\rm{1}}{{\rm{0}}^{ - 16}}\;{\rm{J}} \end{align*} {/eq}

The mass of an electron is {eq}9.11 \times {10^{ - 31}}\;{\rm{kg}} {/eq}.

Substitute the values in equation (I) to calculate the velocity of the electrons.

{eq}\begin{align*} {\rm{29}}{\rm{.12}} \times {\rm{1}}{{\rm{0}}^{ - 16}}\;{\rm{J}} &= \dfrac{1}{2} \times 9.11 \times {10^{ - 31}}\;{\rm{kg}} \times {v^2}\\ {v^2} &= \dfrac{{{\rm{29}}{\rm{.12}} \times {\rm{1}}{{\rm{0}}^{ - 16}}\;{\rm{kg}}{{\rm{m}}^2}{{\rm{s}}^{ - 2}} \times {\rm{2}}}}{{9.11 \times {{10}^{ - 31}}\;{\rm{kg}}}}\\ v &= \sqrt {63.92 \times {{10}^{14}}\;{{\rm{m}}^2}{{\rm{s}}^{ - 2}}} \\ v &= 8 \times {10^7}\;{\rm{m/s}} \end{align*} {/eq}

The wavelength of an electron can be calculated as shown below.

{eq}\lambda = \dfrac{h}{{mv}} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( {{\rm{II}}} \right) {/eq}

Where;

• {eq}\lambda {/eq} is the de Broglie wavelength.
• {eq}m {/eq} is the mass of the electron.
• {eq}v {/eq} is the velocity of the electron
• {eq}h {/eq} is the Planck's constant {eq}\left( {6.626 \times {{10}^{ - 34}}{\rm{J}} \cdot {\rm{s}}} \right) {/eq}

Substitute the values in equation (II) to calculate the wavelength of an electron.

{eq}\begin{align*} \lambda &= \dfrac{{6.626 \times {{10}^{ - 34}}{\rm{J}} \cdot {\rm{s}}}}{{9.11 \times {{10}^{ - 31}}\;{\rm{kg}} \times 8 \times {{10}^7}\;{\rm{m/s}}}}\\ &= \dfrac{{6.626 \times {{10}^{ - 34}}{\rm{kg}}{{\rm{m}}^2}{{\rm{s}}^{ - 2}} \cdot {\rm{s}}}}{{9.11 \times {{10}^{ - 31}}\;{\rm{kg}} \times 8 \times {{10}^7}\;{\rm{m/s}}}}\\ &= 9.09 \times {10^{ - 12}}\;{\rm{m}} \end{align*} {/eq}

Therefore, the wavelength of an electron is {eq}\boxed{{\rm{9.09 \times {10^{ - 12}}\;{\rm{m}}}}} {/eq}