# To resolve an object in an electron microscope, the wavelength of the electrons must be close to...

## Question:

To resolve an object in an electron microscope, the wavelength of the electrons must be close to the diameter of the object. What kinetic energy must the electrons have in order to resolve a protein molecule that is {eq}\rm 2.10\ nm {/eq} in diameter? Take the mass of an electron to be {eq}\rm 9.11 \times 10^{-31}\ kg {/eq}.

## Kinetic energy:

Emitted electrons exhibit motion due to the kinetic energy. This energy is determined if the values of electron's velocity and mass are known. De Broglie equation lets us calculate the value of wavelength.

Given Data:

• The wavelength of protein molecule is 2.10 nm.
• The mass of an electron is {eq}9.11 \times {10^{ - 31}}\;{\rm{kg}} {/eq}.

The velocity 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{I}} \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}

The wavelength is in nanometers.

The conversion of nanometers into meters is done as shown below.

{eq}1\;{\rm{nm}} = {\rm{1}} \times {\rm{1}}{{\rm{0}}^{ - 9}}\;{\rm{m}} {/eq}

The conversion of 2.10 nm into meters is done as shown below.

{eq}2.10\;{\rm{nm}} = 2.10 \times {\rm{1}}{{\rm{0}}^{ - 9}}\;{\rm{m}} {/eq}

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

{eq}\begin{align*} 2.10 \times {\rm{1}}{{\rm{0}}^{ - 9}}\;{\rm{m}} &= \dfrac{{6.626 \times {{10}^{ - 34}}{\rm{J}} \cdot {\rm{s}}}}{{9.11 \times {{10}^{ - 31}}\;{\rm{kg}} \times v}}\\ v &= \dfrac{{6.626 \times {{10}^{ - 34}}{\rm{kg}}{{\rm{m}}^2}{{\rm{s}}^{ - 2}} \cdot {\rm{s}}}}{{2.10 \times {\rm{1}}{{\rm{0}}^{ - 9}}\;{\rm{m}} \times 9.11 \times {{10}^{ - 31}}\;{\rm{kg}}}}\\ v &= 3.46 \times {10^5}\;{\rm{m/s}} \end{align*} {/eq}

The kinetic energy 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{II}}} \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.

Substitute the values in equation (II) to calculate the kinetic energy of electrons.

{eq}\begin{align*} K.E &= \dfrac{1}{2} \times 9.11 \times {10^{ - 31}}\;{\rm{kg}} \times {\left( {3.46 \times {{10}^5}\;{\rm{m/s}}} \right)^2}\\ &= \dfrac{{1.09 \times {{10}^{ - 19}}}}{2}\;{\rm{kg}}{{\rm{m}}^2}{{\rm{s}}^{ - 2}}\\ &= 5.45 \times {10^{ - 20}}\;{\rm{J}} \end{align*} {/eq}

Therefore, the kinetic energy of the electrons is {eq}\boxed{{\rm{5.45 \times {10^{ - 20}}\;{\rm{J}}}}} {/eq} 