Menu Top


Line Spectra and De Broglie's Explanation

The Line Spectra Of The Hydrogen Atom (Calculation of Wavelengths)

As discussed earlier (Section I2 of Rutherford's Model and Atomic Spectra), the emission and absorption spectra of hydrogen consist of discrete lines at specific wavelengths. Bohr's model successfully explained the origin of these lines based on electron transitions between quantised energy levels.

Energy Levels of Hydrogen

According to Bohr's model, the allowed energy levels of the electron in a hydrogen atom ($Z=1$) are given by:

$ E_n = -\frac{13.6}{n^2} \, eV $

Where $n=1, 2, 3, ...$ are the principal quantum numbers. $E_1$ is the energy of the ground state, $E_2$ is the energy of the first excited state, and so on.

Calculation of Wavelengths of Spectral Lines

According to Bohr's third postulate, a spectral line is emitted when an electron makes a transition from a higher energy level $E_{n_i}$ (corresponding to quantum number $n_i$) to a lower energy level $E_{n_f}$ (corresponding to quantum number $n_f$), where $n_i > n_f$. The energy of the emitted photon ($h\nu$) is equal to the difference in the energies of the initial and final levels:

$ h\nu = E_{n_i} - E_{n_f} $

Substitute the energy formula $E_n = -\frac{13.6}{n^2}$:

$ h\nu = \left(-\frac{13.6}{n_i^2}\right) - \left(-\frac{13.6}{n_f^2}\right) $

$ h\nu = 13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \, eV $

The frequency $\nu$ is related to the wavelength $\lambda$ by $c = \nu\lambda$, so $\nu = c/\lambda$.

$ \frac{hc}{\lambda} = 13.6 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \, eV $

Rearranging to find the reciprocal of the wavelength (wavenumber, $1/\lambda$):

$ \frac{1}{\lambda} = \frac{13.6 \, eV}{hc} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) $

The term $\frac{13.6 \, eV}{hc}$ is a constant. Let's calculate its value: $13.6 \, eV = 13.6 \times (1.602 \times 10^{-19} \, J)$. $h = 6.626 \times 10^{-34} \, J \cdot s$. $c = 3 \times 10^8 \, m/s$. $hc = (6.626 \times 10^{-34} \, J \cdot s) \times (3 \times 10^8 \, m/s) = 1.9878 \times 10^{-25} \, J \cdot m$. $ \frac{13.6 \, eV}{hc} = \frac{13.6 \times 1.602 \times 10^{-19} \, J}{1.9878 \times 10^{-25} \, J \cdot m} \approx \frac{21.787 \times 10^{-19}}{1.9878 \times 10^{-25}} \, m^{-1} $ $ \approx 10.96 \times 10^6 \, m^{-1} = 1.096 \times 10^7 \, m^{-1} $

This constant is the Rydberg constant for hydrogen ($R_H$) that appeared in the empirical Rydberg formula for hydrogen spectrum.

$ \frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) $

This formula, derived from Bohr's model, is identical to the empirical formula, with $n_1 = n_f$ and $n_2 = n_i$. This was a major success for Bohr's theory, as it explained the origin of the spectral lines and calculated the Rydberg constant from fundamental physical constants ($R_H = \frac{m_e e^4}{8\epsilon_0^2 h^3 c}$).

The different spectral series correspond to electron transitions ending in different final energy levels ($n_f$):

Diagram showing the hydrogen energy levels and transitions corresponding to spectral series

Energy level diagram showing transitions for hydrogen spectral series.

Example 1. Calculate the wavelength of the H-beta line in the Balmer series of the hydrogen spectrum. (Rydberg constant $R_H = 1.097 \times 10^7 \, m^{-1}$).

Answer:

Given:

Rydberg constant, $R_H = 1.097 \times 10^7 \, m^{-1}$

For the Balmer series, the final state is $n_f = 2$. The lines in the Balmer series correspond to transitions from $n_i = 3, 4, 5, ...$. The first line (H-alpha) is $n_i=3 \to n_f=2$. The second line (H-beta) is $n_i=4 \to n_f=2$.

We use the Rydberg formula: $ \frac{1}{\lambda} = R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) $

Substitute $n_f = 2$ and $n_i = 4$:

$ \frac{1}{\lambda} = (1.097 \times 10^7 \, m^{-1}) \left(\frac{1}{2^2} - \frac{1}{4^2}\right) $

$ \frac{1}{\lambda} = (1.097 \times 10^7) \left(\frac{1}{4} - \frac{1}{16}\right) \, m^{-1} $

$ \frac{1}{\lambda} = (1.097 \times 10^7) \left(\frac{4 - 1}{16}\right) \, m^{-1} $

$ \frac{1}{\lambda} = (1.097 \times 10^7) \left(\frac{3}{16}\right) \, m^{-1} $

$ \frac{1}{\lambda} = \frac{3.291}{16} \times 10^7 \, m^{-1} \approx 0.20569 \times 10^7 \, m^{-1} $

$ \frac{1}{\lambda} \approx 2.0569 \times 10^6 \, m^{-1} $

Now, find the wavelength $\lambda = 1 / (1/\lambda)$:

$ \lambda = \frac{1}{2.0569 \times 10^6 \, m^{-1}} \approx 0.4862 \times 10^{-6} \, m $

$ \lambda = 4.862 \times 10^{-7} \, m = 486.2 \, nm $

The wavelength of the H-beta line in the Balmer series is approximately 486.2 nm. This is in the blue-green part of the visible spectrum.


De Broglie’S Explanation Of Bohr’S Second Postulate Of Quantisation ($ L = n\hbar $)

Bohr's second postulate, $m_e v_n r_n = n\hbar$, was a key assumption that led to the quantisation of energy levels, but it lacked a theoretical explanation within the framework of Bohr's model itself. The quantisation condition seemed arbitrary. With the advent of de Broglie's hypothesis of matter waves, a natural explanation for this postulate emerged.

Electrons as Waves in Orbits

De Broglie proposed that if electrons behave as waves, then these waves must exist as standing waves in the allowed orbits of the atom. A standing wave is a wave that oscillates in time but whose peak amplitude profile does not move in space. Standing waves can only exist for certain wavelengths determined by the boundary conditions of the system (e.g., standing waves on a string fixed at both ends must have wavelengths such that an integral number of half-wavelengths fit within the length of the string).

For an electron wave in a circular orbit around the nucleus, the standing wave condition is that the circumference of the orbit must be an integral multiple of the de Broglie wavelength ($\lambda$) of the electron. If the wave does not "fit" into the orbit in this way, the wave would interfere destructively with itself over successive orbits, and the wave (and hence the electron) would not persist in that orbit.

Let the radius of the electron's orbit be $r$. The circumference of the orbit is $2\pi r$. According to the standing wave condition:

Circumference = $n \times$ de Broglie wavelength

$ 2\pi r = n \lambda $

Where $n$ is an integer ($n=1, 2, 3, ...$), representing the number of wavelengths that fit into the orbit.

Diagram showing electron wave as a standing wave in a circular orbit

Electron wave as a standing wave in an allowed orbit (here $n=3$). The circumference is an integral number of wavelengths.

Now, substitute the de Broglie wavelength $\lambda = h/p = h/(m_e v)$, where $m_e$ is the electron mass and $v$ is its speed in the orbit:

$ 2\pi r = n \left(\frac{h}{m_e v}\right) $

Rearrange this equation:

$ m_e v (2\pi r) = nh $

$ m_e v r = n \frac{h}{2\pi} $

Recall that $m_e v r$ is the magnitude of the angular momentum ($L$) of the electron in its orbit.

$ L = n \frac{h}{2\pi} = n\hbar $

Conclusion

The de Broglie hypothesis, by treating the electron as a wave, naturally leads to the condition that the angular momentum of the electron in an allowed orbit must be an integral multiple of $\hbar$. This is exactly Bohr's second postulate of quantisation of angular momentum, which was previously introduced as an ad-hoc assumption.

This explanation provided strong support for de Broglie's idea of matter waves and demonstrated that Bohr's quantisation condition could be understood as a consequence of the wave nature of the electron. Only those orbits are allowed where the electron wave forms a stable standing wave pattern.

This unification of the wave-particle duality and atomic structure was a significant step towards the development of a more complete quantum theory of the atom (quantum mechanics), where the electron is described by a wave function rather than a classical particle in orbit. The integer $n$ in the standing wave condition ($2\pi r = n\lambda$) is the same principal quantum number that determines the energy level and radius in Bohr's model.

Example 1. Show that the circumference of the first Bohr orbit of the hydrogen atom is equal to the de Broglie wavelength of the electron in that orbit.

Answer:

For the first Bohr orbit of the hydrogen atom, the principal quantum number is $n=1$.

According to Bohr's second postulate, the angular momentum in the first orbit is $L_1 = m_e v_1 r_1 = 1 \cdot \frac{h}{2\pi} = \hbar$.

$ m_e v_1 r_1 = \frac{h}{2\pi} $

Rearranging this equation, we get:

$ 2\pi r_1 = \frac{h}{m_e v_1} $

According to de Broglie's hypothesis, the wavelength ($\lambda$) of a particle with momentum $p = m_e v$ is $\lambda = h/p = h/(m_e v)$.

So, the de Broglie wavelength of the electron in the first Bohr orbit is $\lambda_1 = h/(m_e v_1)$.

From the rearrangement of Bohr's second postulate for $n=1$, we have $2\pi r_1 = \frac{h}{m_e v_1}$.

Comparing the two expressions, we see that:

$ \text{Circumference of first Bohr orbit} = 2\pi r_1 = \frac{h}{m_e v_1} $

$ \text{De Broglie wavelength in first orbit} = \lambda_1 = \frac{h}{m_e v_1} $

$ 2\pi r_1 = \lambda_1 $

Thus, the circumference of the first Bohr orbit is equal to the de Broglie wavelength of the electron in that orbit. This means that one complete de Broglie wavelength fits exactly into the first Bohr orbit, satisfying the standing wave condition for $n=1$. This provides a physical interpretation for Bohr's quantisation rule based on the wave nature of the electron.