A beam of electrons is incident upon a gas of hydrogen atoms. What minimum speed must the electrons have to cause the emission of light from the \(n=3\) to \(n=2\) transition of hydrogen?

In order to calculate the energy difference between n=3 and n=2 energy levels, we will use the Balmer-Rydberg formula for the energy levels of hydrogen which states: \(E_n = -\cfrac{13.6\,\rightarrow eV}{n^2}\) Where, \(E_n\) is the energy of the nth energy level. First, calculate the energy for \(n=3\) and \(n=2\) levels: \(E_3 = -\cfrac{13.6\,\rightarrow eV}{3^2}\) \(E_3 = -1.51\,\rightarrow eV\) and \(E_2 = -\cfrac{13.6\,\rightarrow eV}{2^2}\) \(E_2 = -3.40\,\rightarrow eV\) Now, calculate the energy difference, \(\Delta E\): \(\Delta E = E_2 - E_3\) \(\Delta E = (-3.40\,\rightarrow eV) - (-1.51\,\rightarrow eV)\) \(\Delta E = 1.89\,\rightarrow eV\) The energy difference between n=3 and n=2 levels of hydrogen is 1.89 eV.

Since we are looking for the minimum speed of electrons necessary to cause the emission of light through hydrogen atom transition, we will assume that all of the electron's kinetic energy is transferred to the hydrogen atom. Therefore, the electron's kinetic energy should be equal to the energy difference between the two energy levels: \(K_e = \Delta E\) \(K_e = 1.89\,\rightarrow eV\)

We can now convert the electron's kinetic energy into its speed. The classical equation for kinetic energy is: \(K_e=\cfrac{1}{2}mv^2\), Where \(m\) is the mass of an electron and \(v\) is its speed. First, let's convert the kinetic energy from electron volts (eV) to Joules (J) using the conversion factor of 1 eV = 1.60218 x \(10^{-19}\) J: \(K_e = 1.89\,\rightarrow eV\times (1.60218 \times 10^{-19}\,\rightarrow J/eV)\) \(K_e = 3.02892 \times 10^{-19}\,\rightarrow J\) Now we can solve for the electron's speed, v: \(v = \sqrt{\cfrac{2K_e}{m}}\) The mass of an electron is \(9.11\times 10^{-31}\) kg. Plugging in the values, we get: \(v = \sqrt{\cfrac{2(3.02892 \times 10^{-19}\,\rightarrow J)}{9.11\times 10^{-31}\,\rightarrow kg}}\) \(v = \sqrt{\cfrac{6.05784 \times 10^{-19}\,\rightarrow J}{9.11\times 10^{-31}\,\rightarrow kg}}\) \(v \approx 1. 44 \times 10^6 \,\rightarrow m/s\) Hence, the minimum speed that electrons must have to cause the emission of light from the \(n=3\) to \(n=2\) transition of hydrogen is approximately \(1.44 \times 10^6\,\rightarrow m/s\).