Question

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?

Short answer

Answer: The minimum speed of electrons required is approximately \(1.44 \times 10^6\,\rightarrow m/s\).

Step-by-step solution

Calculate energy difference between n=3 and n=2 levels 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.

Convert energy difference into kinetic energy

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: \(K_e = \Delta E\) \(K_e = 1.89\,\rightarrow eV\)

Convert electron's kinetic energy into its speed

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 \(1.44 \times 10^6\,\rightarrow m/s\).

Key concepts

Energy Level Transitions

When electrons interact with hydrogen atoms, they can cause the electrons in the atom to jump between different energy levels. These energy level transitions are quantified by the values of the principal quantum number, often denoted as \(n\). For hydrogen, when an electron transitions from a higher energy level, such as \(n=3\), to a lower energy level, like \(n=2\), light is emitted. This emitted light can be observed in the form of spectral lines. Such transitions require an external source of energy, often provided by incoming electrons, which imparts enough energy to shift the atomic electron to a higher energy state. These energy transitions form the basis of understanding electromagnetic interactions at the atomic scale.

Balmer-Rydberg Formula

The Balmer-Rydberg formula is crucial for calculating the energy levels in a hydrogen atom. It allows us to determine the precise energy associated with each level \(n\). The formula is expressed as: \[ E_n = -\frac{13.6}{n^2} \text{ eV} \]where \(E_n\) represents the energy of the \(n\)-th energy level in electron volts (eV). Using this formula, we can calculate the specific energy levels of the hydrogen atom. For example, for \(n=3\), the energy is \(-1.51\, \text{eV}\), while for \(n=2\), it is \(-3.40\, \text{eV}\). The energy difference between these two levels can then be calculated to find out how much energy is emitted as a photon when an electron moves between these states.

Kinetic Energy of Electrons

Kinetic energy is a key concept in understanding how electrons transfer energy to hydrogen atoms. When electrons are accelerated to high speeds, they gain kinetic energy, expressed by the formula:\[ K_e = \frac{1}{2}mv^2 \]where \(m\) is the mass of the electron and \(v\) is its velocity. In our scenario, the electron's kinetic energy must be at least equal to the energy difference between the \(n=3\) and \(n=2\) levels of hydrogen to induce the transition. Converting the energy difference from electron volts to joules (since kinetic energy in physics is generally measured in joules), we use 1 eV = 1.60218 x \(10^{-19}\) J, which allows us to calculate the necessary kinetic energy for the interaction.

Speed of Electrons

The speed of electrons required to cause a specific energy level transition in a hydrogen atom can be derived from their kinetic energy. Using the relationship between kinetic energy and speed,\[ v = \sqrt{\frac{2K_e}{m}} \]we calculate the minimum speed for an electron. With a given kinetic energy of \(3.02892 \times 10^{-19}\) J and knowing the mass of an electron is \(9.11 \times 10^{-31}\, \text{kg}\), we find that the electron must travel at approximately \(1.44 \times 10^6\, \text{m/s}\). This high speed is necessary for the electron to have sufficient energy to facilitate the \(n=3\) to \(n=2\) transition, effectively enabling the emission of light at this energy level.