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Question 40.10: CASE STUDY Comparing a Conventional Microscope to an Electro......

CASE STUDY Comparing a Conventional Microscope to an Electron Microscope

In this example, we compare a conventional microscope that illuminates a sample with an electromagnetic wave to an electron microscope (of either design) that uses an electron wave. Suppose you wish to image a strand of DNA, which has a width of roughly 2 nm. You’d like a high-resolution, greatly magnified image, and you estimate that you require a wavelength of about 0.02 nm, so that there will be roughly 100 pixels or resolution elements across the DNA’s width.

A. Find the energy (in eV) of the electromagnetic wave used to illuminate the DNA in a conventional microscope. In the CHECK and THINK step, discuss the practical limitations of such a device.

B. Find the kinetic energy (in eV) of the electron wave used in an electron microscope. Assume that the electrons are not relativistic. In CHECK and THINK, compare your results to those in part A.

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A.

INTERPRET and ANTICIPATE
Use the estimated wavelength to find the energy of the photons in the electromagnetic wave your microscope would employ.

SOLVE
The energy of the photons depends on their frequency, according to Equation 40.8. Their frequency is related to their wavelength by λf = c (Eq. 34.20). Convert the energy to eV.

\begin{aligned}& E=h f  \quad \quad (40.8)\\& E=h \frac{c}{\lambda}=\left(6.63 \times 10^{-34} J \cdot s \right)\left(\frac{3.00 \times 10^8 m / s }{2 \times 10^{-11} m }\right) \\& E=9.9 \times 10^{-15} J \\& 1  eV =1.60 \times 10^{-19} J \\& E=6 \times 10^4 eV\end{aligned}

CHECK and THINK
As shown in Table 34.2 (page 1101), electromagnetic radiation with a wavelength of 2 \times 10^{-11} m is in the X-ray or gamma-ray band. Such high-energy radiation would pose a threat to the researchers, their microscope, and their sample. Using X-rays is not practical in a high-resolution microscope.

B.

INTERPRET and ANTICIPATE
Find the momentum of the electrons from their wavelength, and find their kinetic energy from their momentum.

SOLVE
Use Equation 40.18 to find the electron momentum.

\begin{aligned}\lambda & =\frac{h}{p}  \quad \quad (40.18)\\p & =\frac{h}{\lambda}\end{aligned}

Write the expression for kinetic energy in terms of the momentum. Substitute values, then convert the result to eV.

\begin{aligned}K & =\frac{1}{2} m_e \nu^2=\frac{\left(m_e \nu\right)^2}{2 m_e} \\ K & =\frac{p^2}{2 m_e}=\frac{1}{2 m_e}\left(\frac{h}{\lambda}\right)^2 \\ K & =\frac{1}{2\left(9.11 \times 10^{-31} kg \right)}\left(\frac{6.63 \times 10^{-34} J \cdot s }{2 \times 10^{-11} m }\right)^2 \\K & =6.0 \times 10^{-16} J =4 \times 10^3 eV\end{aligned}

CHECK and THINK
The energy required by the electron microscope is about 15 times lower than that required by the conventional microscope. It will be easier to power an electron microscope and to control the high-energy electrons with magnetic (or electrostatic) fields.

Although the resolution of an electron microscope is not limited significantly by diffraction, like a conventional microscope, it is subject to spherical aberration, distortion, chromatic aberration, and other imaging defects. So even an electron microscope cannot achieve the resolution of 0.02 nm that we would like here. The best resolution achieved by an electron microscope is typically between 0.1 and 0.5 nm, and its highest useful magnification is typically between 10^4 and 10^5. Figure 40.18 is an image of a bundle of strands DNA taken with a transmission electron microscope. The bar in the bottom left corner is the scale; it represents 20 nm. This amazing image shows six DNA stands wrapped around a seventh strand. The inset shows a close-up of the helical structure, with red arrows pointing to the individual turns of the helix. The width of the DNA is 2 nm, so assuming that the resolution is 0.2 nm, we find that there are about 10 resolution elements across the width of a single strand of DNA. Without such detailed resolution, the helical structure would not be seen, the image would look like a smooth line without any structure.

TABLE 34.2 The electromagnetic spectrum broken into convenient bands. Values are approximate.
Name of band Wavelength
λ (m)
Frequency
f (Hz)
Radio >10^{-2} <10^{11}
Microwave 10^{-4}-1 10^9-10^{13}
Infrared (IR) 10^{-6}-10^{-4} 10^{12}-10^{14}
Visible light 10^{-7}-10^{-6} 10^{14}-10^{15}
Ultraviolet (UV) 10^{-9}-10^{-7} 10^{15}-10^{18}
X-rays 10^{-12}-10^{-9} 10^{17}-10^{20}
Gamma rays <10^{-10} >10^{19}
40.18

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