Problem 48 - Fundamentals of Physics

Question

Light of wavelength \(121.6 \mathrm{nm}\) is emitted by a hydrogen atom. What are the (a) higher quantum number and (b) lower quantum number of the transition producing this emission? (c) What is the name of the series that includes the transition?

Step-by-Step Solution

Answer

(a) 2, (b) 1, (c) Lyman series.

Step 1: Understanding the Problem

We are given a wavelength of light emitted from a hydrogen atom and need to find the higher and lower energy levels (quantum numbers) responsible for this emission. This involves identifying which spectral series the transition belongs to.

Step 2: Identify the Spectral Series

The wavelength of 121.6 nm is associated with the Lyman series in the hydrogen spectrum, where transitions end at the first energy level, n=1.

Step 3: Calculate the Energy of the Photon

Use the formula \( E = \frac{hc}{\lambda} \) to calculate the energy of the photon. Given \( \lambda = 121.6 \) nm (or \( 1.216 \times 10^{-7} \) m), Planck's constant \( h = 6.626 \times 10^{-34} \text{ J s} \), and the speed of light \( c = 3 \times 10^8 \text{ m/s} \), we find \( E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{1.216 \times 10^{-7}} = 1.636 \times 10^{-18} \text{ J} \).

Step 4: Energy of Photon in eV

Convert the energy from joules to electron volts using 1 eV = \( 1.602 \times 10^{-19} \text{ J} \). \[ E = \frac{1.636 \times 10^{-18}}{1.602 \times 10^{-19}} \text{ eV} = 10.2 \text{ eV}. \]

Step 5: Determine the Transition Energy Levels

For the Lyman series, we know transitions end at n=1. The energy difference between levels \( n_i \) (initial) and \( n_f \) (final) for a hydrogen atom is \( E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV} \). Solve \( 10.2 = 13.6 (1 - \frac{1}{n_i^2}) \).

Step 6: Solve for the Higher Quantum Number

Plug in the known values: \( 10.2/13.6 = 1 - \frac{1}{n_i^2} \). Rearranging gives \( \frac{1}{n_i^2} = 1 - 0.75 = 0.25 \), therefore \( n_i = 2 \).

Step 7: Conclusion

The higher quantum number (n_i) is 2, the lower quantum number (n_f) is 1, and the transition belongs to the Lyman series.

Key Concepts

Quantum Numbers

Quantum numbers are essential in describing the unique quantum state of an electron in an atom. When dealing with hydrogen spectral series, we frequently refer to two particular quantum numbers: the principal quantum numbers associated with energy levels. These quantum numbers are represented by the symbol "n."

• The higher quantum number (n_i): Describes the initial energy level of an electron before it transitions; larger values mean higher energy levels.
• The lower quantum number (n_f): Describes the final energy level after the electron has emitted or absorbed light, typically a lower value than n_i.

Quantum numbers help in predicting the electron configuration of atoms, providing insights into electron movements and the subsequent energy emitted or absorbed during transitions.
Understanding quantum numbers is crucial for explaining phenomena such as the emission lines in the hydrogen spectrum.

Lyman Series

The Lyman series is a set of ultraviolet spectral lines in a hydrogen atom's emission spectrum. It refers to the electron transitions that terminate at the n=1 energy level, the lowest possible energy state for the hydrogen atom.

• This series corresponds to transitions where electrons fall from a higher energy level to the ground state (n=1).
• The Lyman series includes lines in the ultraviolet region, meaning they are not visible to the naked eye.

In the exercise, the emitted light had a wavelength of 121.6 nm, identifying it as part of the Lyman series. Transitions in the Lyman series typically involve significant energy changes, which is why they emit light in the ultraviolet range.
The understanding of the Lyman series helps in identifying and categorizing the spectral lines in hydrogen, enabling deeper insights into atomic behaviors and properties.

Photon Energy Calculation

Calculating photon energy is fundamental for understanding energy exchanges occurring during electron transitions in atoms. The exercise involves such a calculation, helping to determine the energy difference involved when an electron moves between quantized energy levels.

• The energy of a photon can be found using the formula: \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the light.
• In the exercise, given \( \lambda = 121.6 \) nm, Planck's constant \( h = 6.626 \times 10^{-34} \text{ J s} \), and \( c = 3 \times 10^8 \text{ m/s} \), the calculated energy \( E \) equals \( 1.636 \times 10^{-18} \text{ J} \).
• To convert this energy from joules to electron volts (the preferred unit for atomic scales), use the conversion: 1 eV = \( 1.602 \times 10^{-19} \text{ J} \), resulting in \( 10.2 \text{ eV} \) for this photon.

Grasping how to compute photon energy allows students to better understand not only atomic transitions but also the relation between different units of energy and their significance in scientific experiments.