Question

An image of a biological sample is to have a resolution of \(5 \mathrm{nm} .\) (a) What is the kinetic energy of a beam of electrons with a de Broglie wavelength of \(5.0 \mathrm{nm} ?\) (b) Through what potential difference should the electrons be accelerated to have this wavelength? (c) Why not just use a light microscope with a wavelength of 5 nm to image the sample?

Short answer

Answer: The kinetic energy required for an electron beam is approximately 9.614 x 10^-19 J, and the potential difference needed is around 6.01 kV. The limitations of using a light microscope with a wavelength of 5 nm include potential damage to the biological sample from high-energy photons, the need for specialized lenses for high-energy photon handling, and limited depth information in the obtained images, making it difficult to get detailed three-dimensional imaging. For this reason, using an electron microscope would be more suitable for imaging biological samples at a resolution of 5 nm.

Step-by-step solution

Calculate the de Broglie wavelength of the electrons

Given that the required resolution is 5 nm, we can start by finding out the de Broglie wavelength of the electron beam, which is also 5 nm. The de Broglie wavelength \(\lambda\) is given by the formula: $$\lambda = \frac{h}{p}$$ where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)) and \(p\) is the momentum of the electron.

Calculate the momentum of the electrons

To calculate the momentum of the electrons, we first need to rearrange the de Broglie wavelength formula to solve for \(p\): $$p = \frac{h}{\lambda}$$ Substituting the values for \(h\) and \(\lambda\) into the formula and solving for \(p\): $$p = \frac{6.626 \times 10^{-34} \, \text{Js}}{5 \times 10^{-9} \, \text{m}} = 1.3252 \times 10^{-25} \, \text{kg m/s}$$

Calculate the kinetic energy of the electrons

Now that we have the momentum of the electrons, we can calculate their kinetic energy. The relationship between momentum (\(p\)), mass (\(m\)) and velocity (\(v\)) is given by: $$p = mv$$ The kinetic energy (\(K\)) can be calculated using the formula: $$K = \frac{1}{2}mv^2$$ Solving for \(v\) in the momentum formula and substituting it into the kinetic energy formula, we get: $$K = \frac{p^2}{2m}$$ Using the mass of the electron ((\(9.11 \times 10^{-31} \, \text{kg}\))) and the calculated momentum: $$K = \frac{(1.3252 \times 10^{-25} \, \text{kg m/s})^2}{2(9.11 \times 10^{-31} \, \text{kg})} \approx 9.614 \times 10^{-19} \, \text{J}$$ So, the kinetic energy of the electron beam is approximately \(9.614 \times 10^{-19} \, \text{J}\).

Calculate the potential difference needed to achieve the calculated kinetic energy

The relationship between kinetic energy and potential difference is given by: $$K = eV$$ where \(K\) is kinetic energy, \(e\) is the electron charge (\(1.6 \times 10^{-19} \, \text{C}\)), and \(V\) is the potential difference. Rearranging for \(V\): $$V = \frac{K}{e}$$ Substituting the values for \(K\) and \(e\): $$V = \frac{9.614 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} \approx 6.01 \, \text{kV}$$ A potential difference of about 6.01 kV is needed.

Discuss the limitations of using a light microscope with a wavelength of 5 nm

Using a light microscope with a wavelength of 5 nm to image the sample is not practical due to several reasons. Firstly, the energy of the light photons with a wavelength of 5 nm would be very high, and this could potentially damage the biological sample. Secondly, optical imaging systems have limitations in their resolving power, and using such short wavelengths of light would require the use of lenses and optics capable of handling the high-energy photons. These specialized lenses are not easily available and are likely to be very expensive. Finally, using a light microscope with a wavelength of 5 nm would provide limited depth information for the sample, making it difficult to get detailed, three-dimensional images. Overall, it would be more practical to use an electron microscope to achieve a resolution of 5 nm for imaging biological samples.