Atomic Spectra Slide

PHY-1109 Physics

Atomic Spectra and Spectral Series

Atomic Spectra: Fig. 1 illustrates An idealized spectrometer, showing how light from an excited rarefied gas or vapor passes through a slit and a prism to disperse the light onto a screen. Fig. 2 shows Some of the principal lines in the emission spectra of hydrogen, helium, and mercury. Fig. 3 demonstrates that The dark lines in the absorption spectrum of an element correspond to bright lines in its emission spectrum (specifically illustrated for sodium vapor).

Spectral Series of Hydrogen: The wavelengths of spectral lines for hydrogen are specified by formulas related to various series:

1. Balmer series: $\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right),n=3,4,5\dots$ (1). Fig. 4 shows the Balmer series of hydrogen, where the ${\text{H}}_{\alpha }$ line is red, the ${\text{H}}_{\beta }$ line is blue, the ${\text{H}}_{\gamma }$ and ${\text{H}}_{\delta }$ lines are violet, and the remaining lines are in the near ultraviolet.

2. Lyman series: (Found in the ultraviolet) $\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right),n=2,3,4\dots$ (2).

3. Infrared series: Three spectral series are found in the infrared:

   • Paschen: $\frac{1}{\lambda }=R\left(\frac{1}{3^2}-\frac{1}{n^2}\right),n=4,5,6\dots$ (3).
   • Brackett: $\frac{1}{\lambda }=R\left(\frac{1}{4^2}-\frac{1}{n^2}\right),n=5,6,7\dots$ (4).
   • Pfund: $\frac{1}{\lambda }=R\left(\frac{1}{5^2}-\frac{1}{n^2}\right),n=6,7,8\dots$ (5).

The quantity $R$ is the Rydberg constant, which has a value of $R=1.097\times 10^7{\text{ m}}^{-1}=0.01097{\text{ nm}}^{-1}$. Fig. 5 displays The spectral series of hydrogen, showing how the wavelengths in each series (Lyman, Balmer, Paschen, Brackett, Pfund) are related by simple formulas.

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The Bohr Atom Model

The Bohr Atom model can be examined by considering the wave behavior of an electron orbiting a hydrogen nucleus.

• The de Broglie wavelength ($\lambda$) of the electron is given by $\lambda =\frac{h}{mv}$ (6).
• The centripetal force ($F_c$) is $F_c=\frac{m{v}^2}{r}$.
• The electrostatic force ($F_e$) is $F_e=\frac{1}{4\pi {ϵ}_o}\frac{e^2}{r^2}$. Fig. 6 illustrates the Force balance in the hydrogen atom.
• The condition for orbit stability is $F_c=F_e$, which means $\frac{m{v}^2}{r}=\frac{1}{4\pi {ϵ}_o}\frac{e^2}{r^2}$.

Solving for velocity ($v$) yields: $v=\frac{e}{\sqrt{4\pi {ϵ}_{o}mr}}(7)$. Substituting $v$ back into the de Broglie wavelength formula (6) results in: $\lambda =\frac{h}{e}\sqrt{\frac{4\pi {ϵ}_{o}r}{m}}(8)$.

When substituting the radius of the hydrogen atom, $r=5.3\times 10^{-11}\text{ m}$, the electron wavelength ($\lambda$) is calculated as $33\times 10^{-11}\text{ m}$. This wavelength is exactly the same as the circumference of the electron orbit, $2\pi r=33\times 10^{-11}\text{ m}$.

Fig. 7 shows that The orbit of the electron in a hydrogen atom corresponds to a complete electron de Broglie wave joined on itself. This condition means that an electron can circle a nucleus only if its orbit contains an integral number ($n$) of de Broglie wavelengths, represented by: $n\lambda =2\pi r_n,n=1,2,3,\dots$. Fig. 8 illustrates Some modes of vibration of a wire loop, where the circumference equals an integral number of wavelengths (2, 4, 8). Fig. 9 illustrates that A fractional number of wavelengths cannot persist because destructive interference will occur.

Combining the stability condition $n\lambda =2\pi r_n$ with the derived $\lambda$ from eq. (8) leads to: $\frac{nh}{e}\sqrt{\frac{4\pi {ϵ}_{o}r}{m}}=2\pi r_n(9)$. Solving for $r_n$ shows that the possible electron orbits are those whose radii are: $r_n=\frac{n^2{h}^2{ϵ}_o}{\pi m{e}^2},n=1,2,3,\dots (10)$. The radius of the innermost orbit ($r_1$) is known as the Bohr radius of the Hydrogen Atom, $a_0=r_1=5.292\times 10^{-11}\text{ m}$. Other radii are given by the formula $r_n=n^2{a}_0$ (11).

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Nuclear Motion and Reduced Mass

The standard development of the hydrogen atom theory often assumes the nucleus remains stationary because the proton mass is 1836 times greater than the electron mass. However, the nuclear mass affects the wavelengths of spectral lines.

If the nucleus has a mass $M$ that is not infinite, both the nucleus and the orbital electron revolve around a common center of mass. Fig. 1 illustrates this motion. The distance relationships from the center of mass ($r_N$ from nucleus, $r_E$ from electron, $r=r_E+r_N$) satisfy $m{r}_E=M{r}_N$ (7).

The total angular momentum of the atom is the sum of the angular momentum of the electron and the nucleus, $m\omega r_E^2+M\omega r_N^2$. The angular velocity ($\omega$) is the same for both particles. According to Bohr’s first postulate, the total angular momentum is quantized: $m\omega r_E^2+M\omega r_N^2=n\hslash (9)$.

This angular momentum equation (9) can be rewritten as: $\left(\frac{mM}{M+m}\right)\omega r^2=n\hslash (12)$. Similarly, the force balance condition (centripetal force on the electron equals electrostatic force) is generalized to account for nuclear motion: $m{\omega }^2{r}_E=\frac{1}{4\pi {ϵ}_o}\frac{e^2}{r^2}(13)$. Substituting the expression for $r_E$ from the center of mass calculation, $r_E=\left(\frac{M}{M+m}\right)r$ (10), results in the generalized force balance equation: $\left(\frac{mM}{M+m}\right){\omega }^{2}r=\frac{1}{4\pi {ϵ}_o}\frac{e^2}{r^2}(14)$.

By defining the reduced mass $m^{\prime }$: $m^{\prime }=\left(\frac{mM}{M+m}\right)(15)$. Equations (12) and (14) become, respectively: $m^{\prime }\omega r^2=n\hslash (16)$ $m^{\prime }{\omega }^{2}r=\frac{1}{4\pi {ϵ}_o}\frac{e^2}{r^2}(17)$.

The resulting expression for the energy levels ($E_n$) of the hydrogen atom, incorporating the reduced mass ($m^{\prime }$), is: $E_n=-\frac{m^{\prime }{e}^4}{8{ϵ}_o^2{h}^2}\left(\frac{1}{n^2}\right)(18)$.

The ratio $\frac{m^{\prime }}{m}$ (which equals $\frac{M}{M+m}$) is $\frac{1836}{1837}=0.99945$. Due to the motion of the nucleus, all energy levels of hydrogen are changed by this fraction, leading to an increase of $0.055$ percent (since the energies are less negative).

This correction also affects the Rydberg constant ($R$). Without correcting for nuclear motion, $R$ is $1.0973731\times 10^7{\text{ m}}^{-1}$ (to eight significant figures); the correction lowers it to $1.0967758\times 10^7{\text{ m}}^{-1}$.

Deuterium Identification: The greater nuclear mass of deuterium causes its spectral lines to be shifted slightly to wavelengths shorter than those of ordinary hydrogen. For example, the ${\text{H}}_{\alpha }$ line (a transition from $n=3$ to $n=2$) occurs at $6561\text{ A}$ for deuterium, compared to $6563\text{ A}$ for hydrogen. This slight difference was crucial for the identification of deuterium in 1932 by Harold Urey.

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Hydrogenic Atoms

For a hydrogenic atom (an atom of atomic number $Z$ with only one electron), the nucleus carries a charge $+Ze$. The electrostatic force exerted on the orbital electron is: $F_e=\frac{1}{4\pi {ϵ}_o}\frac{Z{e}^2}{r^2}(19)$. The formula for the energy levels of a hydrogenic atom is consequently: $E_n=-\frac{m^{\prime }{Z}^2{e}^4}{8{ϵ}_o^2{h}^2}\left(\frac{1}{n^2}\right)(20)$. The general formula for the spectral lines ($\frac{1}{\lambda }$) of a hydrogenic atom of atomic number $Z$ is: $\frac{1}{\lambda }=\frac{m^{\prime }{Z}^2{e}^4}{8{ϵ}_o^{2}c{h}^3}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)(21)$. Fig. 2 illustrates the Energy levels in hydrogen and singly ionized helium (${\text{He}}^{+}$).

Related PYQs (Topic III.C, III.D)

• (Topic III.C - Q2) State the postulates of Bohr atom model. Obtain the expressions for the radius and electron energy levels in the $n^{th}$ orbit for $H_2$ atom (2024).
• (Implied Answer) Identification of Deuterium.

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Summary and Application of Uncertainty Principle

The Heisenberg uncertainty principle states that it is impossible to simultaneously measure the x-components of position and of momentum of a particle with an arbitrarily high precision. The product of experimental uncertainties is always larger than or equal to $\frac{1}{2}\hslash$. The limitations of this principle do not relate to the quality of the experimental apparatus but originate in the wave-like nature of matter.

The energy-time uncertainty principle expresses the experimental observation that a quantum state existing only for a short time cannot have a definite energy.

Application of Uncertainty Principle

1. Lower Limit on Energy for an Electron within a Nucleus: To place a lower limit on the energy an electron must have if it is to be part of a nucleus, the uncertainty principle can be applied given a typical atomic nucleus radius of about $5.0\times 10^{-15}\text{ m}$. $\Delta P\ge \frac{\hslash }{2\Delta x}$ $\ge \frac{1.054\times 10^{-34}J\cdot s}{2\times (5.0\times 10^{-15}m)}$ $\ge 1.1\times 10^{-20}\text{ kg}\cdot \text{m/s}$. The minimum kinetic energy is calculated assuming the electron is relativistic ($T=pc$): $\text{Kinetic Energy }T=pc=1.1\times 10^{-20}\text{ kg}\cdot \text{m/s}\times 3\times 10^8\text{ m/s}$. Since $1\text{ eV}=1.6\times 10^{-19}\text{ J}$, the kinetic energy of an electron must exceed $20\text{ MeV}$ if it is to be inside a nucleus. Experiments demonstrate that electrons emitted by certain unstable nuclei never possess more than a small fraction of this energy, leading to the conclusion that nuclei cannot contain electrons.

2. Minimum Energy Estimate for an Electron in a Hydrogen Atom: To estimate the minimum energy an electron can have in a hydrogen atom, given its radius is $5.3\times 10^{-11}\text{ m}$: $\Delta P\ge \frac{\hslash }{2\Delta x}$ $\ge \frac{1.054\times 10^{-34}J\cdot s}{2\times (5.3\times 10^{-11}m)}$. This yields a momentum $\ge 9.9\times 10^{-25}\text{ kg}\cdot \text{m/s}$. An electron with momentum of this magnitude behaves like a classical particle, and its kinetic energy is calculated using $T=\frac{p^2}{2m}$: $T=\frac{(1.1\times 10^{-20}\text{ kg}\cdot \text{m/s}{)}^2}{2\times (9.1\times 10^{-31}\text{ kg})}=5.4\times 10^{-19}\text{ J}$. This energy is $3.4\text{ eV}$. The actual kinetic energy of an electron in the lowest energy level of a hydrogen atom is $13.6\text{ eV}$.

Question	Year(s) of Appearance
Prove that electron can not stay in the nucleus but within the atom on the basis of uncertainty principle.	2020, 2022
Applying uncertainty principle, show that an electron cannot be the part of a nucleus.	2024

Answers to PYQs

• Electron cannot stay in the nucleus: A typical atomic nucleus has a radius $\Delta x\approx 5.0\times 10^{-15}\text{ m}$. Applying the uncertainty principle, the electron’s minimum kinetic energy must exceed $20\text{ MeV}$. However, experiments show that electrons emitted by unstable nuclei possess only a small fraction of this energy, leading to the conclusion that nuclei cannot contain electrons.
• Electron can stay in the atom: A hydrogen atom has a radius $\Delta x\approx 5.3\times 10^{-11}\text{ m}$. The calculated minimum kinetic energy using the uncertainty principle is $3.4\text{ eV}$. Since the actual kinetic energy of an electron in the lowest energy level is $13.6\text{ eV}$, the required energy condition for confinement is satisfied, meaning the electron can stay within the atom.