A beam of electrons passes through a single slit \(40.0 \mathrm{nm}\) wide. The width of the central fringe of a diffraction pattern formed on a screen $1.0 \mathrm{m}\( away is \)6.2 \mathrm{cm} .$ What is the kinetic energy of the electrons passing through the slit?

For this step, we'll use the single slit diffraction formula to find the wavelength of the electrons: $$ \sin \theta = \frac{m\lambda}{a} $$ where \(\theta\) is the angle between the central maximum and the first minimum, \(m\) is an integer (1 for the first minimum), \(a\) is the slit width (40 nm), and \(\lambda\) is the wavelength. We'll first need to calculate the angle \(\theta\), which we can find using the width of the central fringe (\(6.2\) cm) and the distance to the screen (\(1.0\) m). Using the small angle approximation \(\tan \theta \approx \sin \theta \approx \frac{\text{width}}{2\cdot\text{distance}}\), we have: $$ \theta \approx \frac{6.2 \cdot 10^{-2} \mathrm{m}}{2 \cdot 1.0 \mathrm{m}} = 0.031 \mathrm{rad} $$ Now, we can use the single slit diffraction formula to find the wavelength: $$ \lambda = \frac{a\sin \theta}{m} = \frac{40.0\times10^{-9}\mathrm{m}\times0.031\mathrm{rad}}{1} = 1.24\times10^{-9}\mathrm{m} $$

Now that we know the wavelength of the electrons, we can use the de Broglie wavelength formula to find the momentum of the electrons: $$ \lambda = \frac{h}{p} $$ where \(\lambda\) is the wavelength, \(h\) is the Planck constant (\(6.63\times10^{-34}\mathrm{Js}\)), and \(p\) is the momentum of the electrons. Solving for the momentum \(p\), we have: $$ p = \frac{h}{\lambda} = \frac{6.63\times10^{-34}\mathrm{Js}}{1.24\times10^{-9}\mathrm{m}} = 5.35\times10^{-25}\mathrm{kg\cdot m/s} $$

We now have the momentum \(p\) of the electrons, so we can use the classical formula for kinetic energy to find their energy: $$ K = \frac{p^2}{2m} $$ where \(K\) is the kinetic energy and \(m\) is the mass of an electron (\(9.11\times10^{-31}\mathrm{kg}\)). Plugging in the momentum and mass, we get: $$ K = \frac{(5.35\times10^{-25}\mathrm{kg\cdot m/s})^2}{2\cdot(9.11\times10^{-31}\mathrm{kg})} = 1.58 \times 10^{-17}\mathrm{J} $$ So, the kinetic energy of the electrons passing through the slit is approximately \(1.58 \times 10^{-17}\mathrm{J}\).