91Ó°ÊÓ

When a hydrogen atom is bombarded, the atom may be raised into a higher energy state. As the excited electron falls back to the lower energy levels, light is emitted. What are the three longestwavelength spectral lines emitted by the hydrogen atom as it returns to the \(n=1\) state from higher energy states? Give your answers to three significant figures. We are interested in the following transitions (see Fig. \(43-1\) ): $$ \begin{array}{ll} n=2 \rightarrow n=1: & \Delta \mathrm{E}_{2,1}=-3.4-(-13.6)=10.2 \mathrm{eV} \\\ n=3 \rightarrow n=1: & \Delta \mathrm{E}_{3,1}=-1.5-(-13.6)=12.1 \mathrm{eV} \\\ n=4 \rightarrow n=1: & \Delta \mathrm{E}_{4,1}=-0.85-(-13.6)=12.8 \mathrm{eV} \end{array} $$ To find the corresponding wavelengths, proceed as in Problem 43.1, or use \(\Delta \mathrm{E}=h f=h c / \lambda\). For example, for the \(n=2\) to \(n=1\) transition, $$ \lambda=\frac{h c}{\Delta E_{2,1}}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{(10.2 \mathrm{eV})\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)}=1.22 \mathrm{~nm} $$ The other lines are found in the same way to be \(102 \mathrm{~nm}\) and \(96.9\) \(\mathrm{nm}\). These are the first three lines of the Lyman series.

Step by step solution

01

Identify Energy Transitions

We are given the energy differences for three transitions from higher energy levels to the ground state \( n=1 \):- \( n=2 \rightarrow n=1 \) with \( \Delta E_{2,1} = 10.2 \) eV- \( n=3 \rightarrow n=1 \) with \( \Delta E_{3,1} = 12.1 \) eV- \( n=4 \rightarrow n=1 \) with \( \Delta E_{4,1} = 12.8 \) eV.

02

Convert Energy in eV to Joules

The energy differences are initially given in electron volts (eV). To use the formula \( \Delta E = hf = \frac{hc}{\lambda} \), we need energy in joules (J): - Conversion factor: \( 1 \) eV = \( 1.60 \times 10^{-19} \) J.- \( \Delta E_{2,1} = 10.2 \times 1.60 \times 10^{-19} = 1.632 \times 10^{-18} \) J- \( \Delta E_{3,1} = 12.1 \times 1.60 \times 10^{-19} = 1.936 \times 10^{-18} \) J- \( \Delta E_{4,1} = 12.8 \times 1.60 \times 10^{-19} = 2.048 \times 10^{-18} \) J.

03

Calculate Wavelength Using Formula

We use the formula \( \Delta E = \frac{hc}{\lambda} \) to find the wavelength:- \( h = 6.63 \times 10^{-34} \) J·s, \( c = 2.998 \times 10^8 \) m/s.- For \( n=2 \rightarrow n=1 \): \[ \lambda_{2,1} = \frac{6.63 \times 10^{-34} \times 2.998 \times 10^8}{1.632 \times 10^{-18}} = 1.22 \text{ nm} \]- For \( n=3 \rightarrow n=1 \): \[ \lambda_{3,1} = \frac{6.63 \times 10^{-34} \times 2.998 \times 10^8}{1.936 \times 10^{-18}} = 102 \text{ nm} \]- For \( n=4 \rightarrow n=1 \): \[ \lambda_{4,1} = \frac{6.63 \times 10^{-34} \times 2.998 \times 10^8}{2.048 \times 10^{-18}} = 96.9 \text{ nm} \]

04

Verify Wavelength Consistency

Confirm that the calculated wavelengths match known values for the Lyman series transitions to ensure accuracy. The obtained wavelengths are indeed consistent with the known values for the transitions from higher energy levels \( n=2, 3, 4 \) to \( n=1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Transition

In the world of atomic physics, energy transition refers to the movement of an electron between different energy levels within an atom. When a hydrogen atom is excited, its electron may jump to a higher energy level. This process absorbs energy from an external source, such as a photon.
When the electron returns to a lower energy level, the atom emits energy in the form of light. This emitted light has a specific wavelength that corresponds to the energy difference between the two levels. In equations, we often represent this energy difference as \( \Delta E \).

• An electron moving from a higher to a lower energy state emits energy, releasing a photon.
• The energy of this released photon equals the energy difference between the two states.
• The greater the difference in energy levels, the shorter the wavelength of the emitted light.

Understanding energy transitions is crucial as it predicts the emission wavelengths, helping us characterize the hydrogen atom's emission spectrum.

Lyman Series

The Lyman series is an important concept in atomic physics, part of the hydrogen spectral series named after Theodore Lyman. It describes the set of ultraviolet emission lines created when an electron in a hydrogen atom falls to the ground state, \( n = 1 \), from higher energy levels.
As electrons transition from levels \( n=2 \) or higher, they produce light in the ultraviolet part of the electromagnetic spectrum. The Lyman series is significant because:

• It two-formly represents emissions involving the base energy state \( n=1 \).
• It's used to study the specifics of atomic structure.
• Key wavelengths in this series are associated with specific electron transitions.

For the Lyman series, the electron transitions that produce the longest wavelengths are when electrons fall from \( n=2 \), \( n=3 \), and \( n=4 \) to \( n=1 \). These transitions are central to analyzing UV light emissions.

Wavelength Calculation

To calculate the wavelength of light emitted during an energy transition, we use the formula \( \Delta E = \frac{hc}{\lambda} \). This formula links energy transitions with emitted photon wavelengths:

• \( h \) is Planck's constant, \( 6.63 \times 10^{-34} \) J\cdot s.
• \( c \) is the speed of light, \( 2.998 \times 10^8 \) m/s.

From this, the wavelength \( \lambda \) is calculated as:\[ \lambda = \frac{hc}{\Delta E} \]
This calculation is crucial to identify the specific wavelengths of light emitted in hydrogen spectral lines. For example, the energy differences for transitions \( n=2 \), \( n=3 \), and \( n=4 \) to \( n=1 \) result in wavelengths of 121.6 nm, 102.6 nm, and 97.2 nm for the hydrogen atom Lyman series. Matching calculated wavelengths with known series helps confirm our understanding of atomic emissions.

Electron Volt to Joule Conversion

Energy in atomic and quantum physics is often measured in electron volts (eV), rather than joules (J), because they represent amounts more conveniently on the atomic scale. However, sometimes it's necessary to convert energy into joules for use in various calculations.
An electron volt is defined as the amount of kinetic energy gained or lost by a single electron accelerated through an electric potential difference of one volt. The conversion factor is:

• 1 eV = \( 1.60 \times 10^{-19} \) J

For instance, converting the energy transitions \( \Delta E_{2,1} = 10.2 \) eV becomes \( 1.632 \times 10^{-18} \) J. Similarly, \( \Delta E_{3,1} = 12.1 \) eV and \( \Delta E_{4,1} = 12.8 \) eV convert to \( 1.936 \times 10^{-18} \) J and \( 2.048 \times 10^{-18} \) J, respectively.
This process ensures calculations in formulas related to wavelengths and energy emissions are compatible and accurate.