91影视

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 鈥渟ee鈥� 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,

$位=10脳{10}^{鈭�3}\text{鈥塶尘}$

The kinetic energy for electron microscope is,

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

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

The energy for a light microscope is,

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

Take the value of $hc$as,

$hc=1240\text{鈥塭痴}鈰�\text{nm}$

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,

$\begin{array}{rcl}K& =& \sqrt{{\left(\frac{hc}{位}\right)}^2+m_e^2{c}^4}鈭�m_e{c}^2\\ & =& \sqrt{\left(\frac{1240\text{eV}鈰�\text{nm}}{10脳{10}^{鈭�3}\text{nm}}\right)+{\left(0.511\text{鈥塎别痴}\right)}^2}鈭�0.511\text{鈥塎别痴}\\ & =& 0.015\text{鈥塎别痴}\left(\frac{1000\text{鈥�}keV}{1\text{鈥�}MeV}\right)\\ & =& 15\text{鈥塳别痴}\end{array}$

Hence, the minimum energy required by a photon under an electron microscope is $\begin{array}{rcl}& & 15\text{鈥塳别痴}\end{array}$.

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

$E=\frac{hc}{位}$

Substitute known values in the above equation.

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

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

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