Question

Calculate the frequency (Hz) and wavelength (nm) of the emitted photon when an electron drops from the \(n=4\) to the \(n=3\) level in a hydrogen atom.

Short answer

Frequency: approximately 1.595 × 10¹⁴ Hz; Wavelength: approximately 1880 nm.

Step-by-step solution

Calculate the Energy Difference

To calculate the frequency and wavelength of the emitted photon, we must first find the energy difference between the two energy levels. In a hydrogen atom, the energy of an electron is given by the formula: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \]For the \( n=4 \) level:\[ E_4 = -\frac{13.6}{4^2} \, \text{eV} = -\frac{13.6}{16} \, \text{eV} = -0.85 \, \text{eV} \]For the \( n=3 \) level:\[ E_3 = -\frac{13.6}{3^2} \, \text{eV} = -\frac{13.6}{9} \, \text{eV} = -1.51 \, \text{eV} \]The energy difference \( \Delta E \) is:\[ \Delta E = E_3 - E_4 = (-1.51) - (-0.85) \, \text{eV} = -0.66 \, \text{eV} = 0.66 \, \text{eV} \]

Convert Energy to Frequency

The frequency \( u \) of the emitted photon can be related to the energy difference \( \Delta E \) using Planck's equation:\[ \Delta E = h u \]Where \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \) is Planck's constant. First, we need to convert \( \Delta E \) from electronvolts to joules:\[ 0.66 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 1.05732 \times 10^{-19} \, \text{J} \]Now, solve for \( u \):\[ u = \frac{\Delta E}{h} = \frac{1.05732 \times 10^{-19}}{6.626 \times 10^{-34}} \, \text{Hz} \]\[ u \approx 1.595 \times 10^{14} \, \text{Hz} \]

Convert Frequency to Wavelength

The relationship between the frequency \( u \) and the wavelength \( \lambda \) of the light is given by the speed of light \( c \):\[ c = \lambda u \]Where \( c = 3 \times 10^8 \ \, \text{m/s} \). Solve for \( \lambda \):\[ \lambda = \frac{c}{u} = \frac{3 \times 10^8}{1.595 \times 10^{14}} \, \text{m} \]\[ \lambda \approx 1.88 \times 10^{-6} \, \text{m} \]Convert meters to nanometers:\[ 1.88 \times 10^{-6} \, \text{m} \times 10^9 \, \text{nm/m} = 1880 \, \text{nm} \]

Key concepts

Energy Levels

When discussing hydrogen atoms, it's essential to understand the concept of energy levels. In basic terms, electrons in a hydrogen atom orbit the nucleus at specific, quantized energy levels. These levels are designated by quantum numbers, usually denoted as \( n \). Each level represents a different state of energy for the electron.
The formula to determine the energy \( E_n \) of an electron at a particular level \( n \) is given by:

• \( E_n = - \frac{13.6 \, \text{eV}}{n^2} \)

The constant \( 13.6 \, \text{eV} \) represents the energy of the electron at the ground state (\( n = 1 \)). As \( n \) increases, the absolute value of \( E_n \) decreases, meaning the electron's energy becomes less negative, and it is less tightly bound to the nucleus.
Transitions between these levels involve absorbing or emitting energy in the form of photons, which are packets of electromagnetic radiation. For instance, when an electron transitions from the \( n=4 \) to \( n=3 \) level, the energy difference represents the energy of the photon emitted during this transition.

Planck's Constant

Planck's constant is fundamental in quantum mechanics, playing a crucial role in understanding light and matter interactions. It is denoted by \( h \) and has the value \( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \). This constant relates the energy of a photon to its frequency through the equation:

• \( E = h u \)

Where \( E \) is the energy of the photon, and \( u \) (nu) is its frequency.
The concept can be difficult at first because it implies that energy can be measured in discrete packets instead of being continuous. In our example, to find the frequency of the emitted photon when the electron transitions from \( n=4 \) to \( n=3 \), we used the energy difference \( 0.66 \, \text{eV} \), converted it to joules, and used Planck's equation to solve for \( u \).
Planck's constant is key to understanding phenomena at a quantum level. It underscores how energy transitions at atomic scales follow distinct steps rather than continuous flows.

Photon Emission

Photon emission is a process that occurs when an electron drops to a lower energy level, releasing energy in the form of a photon.
This photon carries both a quantifiable amount of energy and a specific wavelength related to the transition. In quantum mechanics, transitions between energy levels are associated with precise photonic emissions, which contributes to the characteristic spectral lines of different elements.

• The energy of the emitted photon \( \Delta E \) determines its frequency \( u \), calculated by Planck's equation: \( \Delta E = h u \).
• Utilizing the speed of light formula, \( c = \lambda u \), the wavelength \( \lambda \) of the emitted photon can be determined, revealing its position in the electromagnetic spectrum.

For example, the electron dropping from \( n=4 \) to \( n=3 \) in a hydrogen atom emits a photon in the infrared range, producing light with a wavelength calculated in nanometers.
Photon emission is not just a fundamental concept in physics, but also a pivotal mechanism underlying technologies like lasers and LED lighting, making the understanding of these quantum processes valuable in both scientific and technological contexts.