De-broglie's Explanation Of Bohr's Second Postulate

De-Broglie's Explanation of Bohr's Second Postulate

Since Bohr gave many postulates in his theory, the second postulate is not very clear and little puzzling. The Scientist De Broglie explained this puzzle very clearly as to why the angular momentum of the revolving electron is the integral multiple of the $h / 2 \pi$. De Broglie in his experiment proved that the electron revolving in a circular orbit has a wave nature also in the last chapter we saw the experiment performed by Davison and Germer which proved that the electron shows the wave nature. In analogy to waves travelling on a string, particle waves too can lead to standing waves under resonant conditions. During the chapter Waves and Oscillation, we know that when a string is plucked, a vast number of wavelengths are excited. However, only those wavelengths survive which have nodes at the ends and form the standing wave in the string. It means that in a string, standing waves are formed when the total distance travelled by a wave down the string and back is any integral number of wavelengths. Waves with other wavelengths interfere with themselves upon reflection and their amplitudes quickly drop to zero.

For an electron moving in $n^{\text {th }}$ circular orbit of radius $r_n$, the total distance is the circumference of the orbit, $2 \pi r_n$.

$2 \pi r_n=n \lambda, \quad n=1,2,3 \ldots$

The figure given above illustrates a standing particle wave on a circular orbit for n = 4, i.e., 2πr_n = 4λ, where λ is the de Broglie wavelength of the electron moving in n^th orbit. From the last chapter, we have studied that λ = h/p, where p is the magnitude of the electron’s momentum. If the speed of the electron is much less than the speed of light, the momentum is mv_n.

Thus,

$
\lambda=\frac{h}{m v_n}
$

From the above equation, we have,
$
2 \pi r_n=\frac{n h}{m v_n} \quad \text { or }, \quad m v_n r_n=\frac{n h}{2 \pi}
$

This is the quantum condition proposed by Bohr for the angular momentum of the electron. Thus de Broglie hypothesis provided an explanation for Bohr’s second postulate for the quantisation of angular momentum of the orbiting electron.

The quantised electron orbits and energy states are due to the wave nature of the electron and only resonant standing waves can persist. Bohr’s model, involving a classical trajectory picture (planet-like electron orbiting the nucleus), correctly predicts the gross features of the hydrogenic atoms(Hydrogenic atoms are the atoms consisting of a nucleus with positive charge +Ze and a single electron, where Z is the proton number. Examples are a hydrogen atom, singly ionised helium, doubly ionised lithium, and so forth.), in particular, the frequencies of the radiation emitted or selectively absorbed.

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Solved Examples Based on De-Broglie's Explanation of Bohr's Second Postulate

Example 1: An electron and a photon have the same wavelength. If $p$ is the momentum of the electron and $\mathbf{E}$ is the energy of the photon. The magnitude of $\frac{p}{E}$ in S.I unit is :

1) $3.0 \times 10^8$
2) $3.33 \times 10^{-9}$
3) $9.1 \times 10^{-31}$
4) $6.64 \times 10^{-34}$

Solution:

From DeBroglie relation $$ \lambda=\frac{\mathrm{h}}{\mathrm{p}} $$ and using energy relation $$ \begin{aligned} & E=\frac{h c}{\lambda} \text { or } \\ & \lambda=\frac{h c}{E} \end{aligned} $$ Equating these two $$ \frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}} $$ hence, $\frac{\mathrm{P}}{\mathrm{E}}=\frac{1}{\mathrm{c}}=3.33 * 10^{-9}$

Example 2: When the kinetic energy of an electron is increased, the wavelength of the associated wave will :

1) Increase

2) Decrease

3) Wavelength does not depend on the kinetic energy

4) None of the above

Solution:

De Broglie wavelength is given by :

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m K}} ; \quad \therefore \lambda \propto \frac{1}{\sqrt{K}}(\mathrm{~h}$ and $\mathrm{m}=$ constant $)$

When the kinetic energy of an electron is increased, the wavelength of the associated wave will be decreased.

Hence, the answer is the option (2).

Example 3: Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65 A^0$ ). The de-Broglie wavelength of this electron is:
1) $3.5 . A^0$
2) $6.6 A^0$
3) $12.9 A^0$
4) $9.7 A^0$

Solution:

The angular momentum of an electron in a stationary orbit is quantized.

$\begin{aligned} & 2 \pi r_n=n \lambda_n \\ & \lambda_3=\frac{2 \pi\left(4.65 \times 10^{-10}\right)}{3} \\ & \lambda_3=9.7 \AA\end{aligned}$

Hence the answer is the option (4).

Example 4: The acceleration of an electron in the first orbit of the hydrogen atom (n=1) is:

1) $\frac{h^2}{\pi^2 m^2 r^3}$
2) $\frac{h^2}{8 \pi^2 m^2 r^3}$
3) $\frac{h^2}{4 \pi^2 m^2 r^3}$
4) $\frac{h^2}{4 \pi m^2 r^3}$

Solution:

Bohr quantisation principle

$
m v r=\frac{n h}{2 \pi}
$
wherein
The angular momentum of an electron in a stationary orbit is quantised.
$
\begin{aligned}
& \text { Acceleration }=\frac{v^2}{r} \\
& \because m v r=\frac{n h}{2 \pi} \Rightarrow v=\frac{h}{2 \pi m r}(n=1) \\
& \therefore a=\left(\frac{h}{2 \pi m r}\right)^2 \cdot \frac{1}{r}=\frac{h^2}{4 \pi^2 m^2 r^3}
\end{aligned}
$

Hence, the answer is the option (3).

Example 5: Suppose an electron is attracted towards the origin by a force k / r where k is constant and r is the distance of the electron from the origin. By applying the Bohr model to this system, the radius of the n^th orbital of the electron is found to be r_n and the kinetic energy of the electron to be T_n. Then which of the following is true?

1) $T_n \alpha \frac{1}{n}, r_n \alpha n^2$
2) $T_n \alpha \frac{1}{n^2}, r_n \alpha n^2$
3) $T_n$ Independent of $n, \quad r_n \alpha n$
4) $T_n \alpha \frac{1}{n}, r_n \alpha n$

Solution:

Supposing that the force of attraction in Bohr's atom does not follow inverse square law but is inversely proportional to r,

$\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r}$ would have been $=\frac{m v^2}{r}$
$
\begin{gathered}
\therefore m v^2=\frac{e^2}{4 \pi \varepsilon_0} \\
\Rightarrow T_n=\frac{1}{2} m \nu^2=\frac{1}{2}\left(\frac{e^2}{4 \pi \varepsilon_0}\right)
\end{gathered}
$

Thus $T_n$ is independent of $n$
From $m \nu r_n=\frac{n h}{2 \pi}$

$
\begin{aligned}
& \text { as } m v^2=\frac{e^2}{4 \pi \varepsilon_0} \\
& m^2 v^2=\frac{e^2 m}{4 \pi \varepsilon_0} \\
& \therefore m v=\sqrt{\frac{e^2 m}{4 \pi \varepsilon_0}}
\end{aligned}
$
$m v$ is independent of $n$,
By Bohr's quantization principle for angular momentum
$
\begin{aligned}
& m \nu r_n=\frac{n h}{2 \pi} \\
& \therefore r_n \propto n
\end{aligned}
$

Hence, the answer is the option (3).

Summary

De Broglie's explanation of Bohr's second postulate sheds light on the quantization of electron orbits by introducing the concept of wave-particle duality. De Broglie proposed that electrons exhibit wave-like properties, leading to the formation of standing waves at specific energy levels. This idea aligns with Bohr's quantized angular momentum and explains the stability of electron orbits. In practical terms, this theory underpins technologies like electron microscopy and semiconductor devices, where understanding electron behaviour at quantum levels is essential for advancements in materials science and electronics.