Question

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of \(9.38 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

Short answer

The characteristic wavelength of the electron is approximately \(8.03 \times 10^{-11}~ \mathrm{m}\). This wavelength is within the size range of atoms, which is typically between \(1 \times 10^{-10}~ \mathrm{m}\) to \(5 \times 10^{-10}~ \mathrm{m}\). Therefore, the electron microscope can resolve individual atoms, although its resolution will depend on factors such as the specific atom's size and the microscope's capabilities.

Step-by-step solution

Understand the de Broglie wavelength formula

The de Broglie wavelength formula describes the wavelength of a particle as a function of its momentum. It is given by the equation: \[ \lambda = \frac{h}{p} \] where \(\lambda\) is the wavelength, \(h\) is the Planck's constant (\(6.63 \times 10^{-34}~ \mathrm{Js}\)), and \(p\) is the momentum of the particle. Since momentum is the product of mass and velocity (\(p = mv\)), we can rewrite the equation as: \[ \lambda = \frac{h}{mv} \] We are given the velocity (\(v = 9.38 \times 10^6~ \mathrm{m/s}\)) and we know the mass of the electron (\(m_e = 9.11 \times 10^{-31}~ \mathrm{kg}\)). With this information, we can find the characteristic wavelength of the electron.

Calculate the wavelength using the given velocity

Plug in the given values of the velocity and mass of the electron into the de Broglie wavelength equation: \[ \lambda = \frac{6.63 \times 10^{-34}~ \mathrm{Js}}{(9.11 \times 10^{-31}~ \mathrm{kg})(9.38 \times 10^{6}~ \mathrm{m/s})} \] After plugging in the values, perform the calculations: \[ \lambda \approx 8.03 \times 10^{-11}~ \mathrm{m} \]

Compare the wavelength to the size of atoms

The wavelength we found, \(\lambda \approx 8.03 \times 10^{-11}~ \mathrm{m}\), is in the picometer range. The size of atoms is typically in the range of 0.1 to 0.5 nanometers, or \(1 \times 10^{-10}~ \mathrm{m}\) to \(5 \times 10^{-10}~ \mathrm{m}\). The electron's wavelength (\(8.03 \times 10^{-11}~ \mathrm{m}\)) is within the size range of atoms, specifically around the lower end of the range. The electron microscope can thus resolve individual atoms, but its resolution will depend on the specific atom's size and other factors, such as the microscope's capabilities and imaging conditions.