Question

The highest achievable resolving power of a microscope is limited only by the wavelength used; that is, the smallest item that can be distinguished has dimensions about equal to the wavelength. Suppose one wishes to “see” inside an atom. Assuming the atom to have a diameter of $\mathbf{\text{100pm}}$, this means that one must be able to resolve a width of, say, $\mathbf{10}\mathbf{\text{pm}}$.

(a) If an electron microscope is used, what minimum photon energy is required?

(b) If a light microscope is used, what minimum photon energy is required?

(c) Which microscope seems more practical? Why?

Short answer

(a) The minimum photon energy required is $15\text{keV}$.

(b) The minimum photon energy required is $120\text{keV}$.

(c) An electron microscope is more suitable.

Step-by-step solution

The given data:

The highest achievable resolving power of a microscope is limited only by the wavelength used; that is, the smallest item that can be distinguished has dimensions about equal to the wavelength.

Suppose one wishes to “see” inside an atom.

The atom has a diameter of $100\text{pm}$, this means that one must be able to resolve a width of, say, $10\text{pm}$.

Thus wavelength is,

$\lambda =10\times {10}^{-3}\text{\hspace{0.17em}nm}$

Concept and Formula used:

The kinetic energy for electron microscope is,

$\mathit{K}\mathbf{=}\sqrt{{\left(\frac{hc}{\lambda }\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{e}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{4}}}\mathbf{-}{\mathit{m}}_{\mathbf{e}}{\mathit{c}}^{\mathbf{2}}$

Here, $\mathit{h}$ is Plank’s constant, $\mathit{c}$is the speed of light, $\mathit{\lambda }$is wavelength, ${\mathit{m}}_{\mathbf{e}}$is the mass of the electron.

The energy for a light microscope is,

$\mathit{E}\mathbf{=}\frac{\mathbf{hc}}{\mathbf{\lambda }}$

(a) Find energy required by photon under the electron microscope:

Take the value of $hc$as,

$hc=1240\text{\hspace{0.17em}eV}\cdot \text{nm}$

And take the value of $m_e{c}^2$as,

$m_e{c}^2=0.511\text{MeV}$

The kinetic energy K for electron microscope is,

$\begin{array}{rcl}K& =& \sqrt{{\left(\frac{hc}{\lambda }\right)}^2+m_e^2{c}^4}-m_e{c}^2\\ & =& \sqrt{\left(\frac{1240\text{eV}\cdot \text{nm}}{10\times {10}^{-3}\text{nm}}\right)+{\left(0.511\text{\hspace{0.17em}MeV}\right)}^2}-0.511\text{\hspace{0.17em}MeV}\\ & =& 0.015\text{\hspace{0.17em}MeV}\left(\frac{1000keV}{1MeV}\right)\\ & =& 15\text{\hspace{0.17em}keV}\end{array}$

Hence, the minimum energy required by a photon under an electron microscope is $\begin{array}{rcl}& & 15\text{\hspace{0.17em}keV}\end{array}$.

(b) Find energy required by photon under photon microscope:

The energy for the light microscope required by the photon is,

$E=\frac{hc}{\lambda }$

Substitute known values in the above equation.

role="math" localid="1663144156968" $\begin{array}{rcl}E& =& \frac{1240\text{eV}\cdot \text{nm}}{10\times {10}^{-3}\text{nm}}\\ & =& 1.2\times {10}^5\text{eV}\left(\frac{1keV}{{10}^{3}eV}\right)\\ & =& 120\text{keV}\end{array}$

Hence, the minimum energy required by a photon under a light microscope is $\begin{array}{rcl}& & 120\text{keV}\end{array}$.

(c) Compare the microscopes:

The electron microscope is more suitable, as the required energy on the electrons is much less than that of the photons.