2.5 Towards Quantum Mechanical Model of the Atom

In view of the shortcoming of the Bohr’s model, scientists attempted to develop a more suitable and general model for atoms. Two important developments which contributed significantly in the formulation of such a model were:

1. Dual behaviour of matter,

2. Heisenberg uncertainty principle.

Dual behaviour of matter

The French physicist, de Broglie, in 1924 proposed that matter, like radiation, should also exhibit dual behaviour i.e., both particle and wavelike properties. This means that just as the photon has momentum as well as wavelength, electrons should also have momentum as well as wavelength, de Broglie, from this analogy, gave the following relation between wavelength \((\lambda)\) and momentum (p) of a material particle.

\(\lambda=\frac{h}{m{v}}=\frac{h}{p} \dots …..(2.22)\)

where \(m\) is the mass of the particle, \(v\) its velocity and \(p\) its momentum. de Broglie’s prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction, a phenomenon characteristic of waves. This fact has been put to use in making an electron microscope, which is based on the wavelike behaviour of electrons just as an ordinary microscope utilises the wave nature of light. An electron microscope is a powerful tool in modern scientific research because it achieves a magnification of about 15 million times.

Problem 2.12
What will be the wavelength of a ball of mass \(0.1 \mathrm{~kg}\) moving with a velocity of 10 \(\mathrm{m}  \mathrm{s}^{-1} ?\)

Solution:
According to de Brogile equation
\(
\begin{aligned}
&\lambda=\frac{h}{m {v}}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)}{(0.1 \mathrm{~kg})\left(10 \mathrm{~m} \mathrm{~s}^{-1}\right)} \\
&=6.626 \times 10^{-34} \mathrm{~m}\left(\mathrm{~J}=\mathrm{kg} \mathrm{m}^{2} \mathrm{~s}^{-2}\right)
\end{aligned}
\)

Problem 2.13
The mass of an electron is \(9.1 \times 10^{-31} \mathrm{~kg}\). If its K.E. is \(3.0 \times 10^{-25} \mathrm{~J}\), calculate its wavelength.

Solution:
Since K. E. \(=1 / 2 \mathrm{~m} \mathrm{v}^{2}\)
\(\mathrm{v}=\left(\frac{2 \mathrm{~K} . \mathrm{E} .}{\mathrm{m}}\right)^{1 / 2}=\left(\frac{2 \times 3.0 \times 10^{-25} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}}{9.1 \times 10^{-31} \mathrm{~kg}}\right)^{1 / 2}\)
\(=812 \mathrm{~m} \mathrm{~s}^{-1}\)
\(\lambda=\frac{h}{m \mathrm{v}}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(812 \mathrm{~ms}^{-1}\right)}\)
\(=8967 \times 10^{-10} \mathrm{~m}=896.7 \mathrm{~nm}\)

Problem 2.14
Calculate the mass of a photon with wavelength \(3.6 Å\).

Solution:
\(\lambda=3.6 Å=3.6 \times 10^{-10} \mathrm{~m}\)
Velocity of photon = velocity of light
\(m=\frac{h}{\lambda \mathrm{v}}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(3.6 \times 10^{-10} \mathrm{~m}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}\)
\(=6.135 \times 10^{-29} \mathrm{~kg}\)

Heisenberg’s Uncertainty Principle

Werner Heisenberg a German physicist in 1927, stated the uncertainty principle which is the consequence of dual behaviour of matter and radiation. It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.

Mathematically, it can be given as in equation (2.23).

\(
\begin{aligned}
&\Delta \mathrm{x} \times \Delta p_{\mathrm{x}} \geq \frac{h}{4 \pi} \dots …..(2.23)\\
&\text { or } \Delta \mathrm{x} \times \Delta\left(m{v}\right) \geq \frac{h}{4 \pi} \\
&\text { or } \Delta \mathrm{x} \times \Delta \mathrm{v}_{\mathrm{x}} \geq \frac{h}{4 \pi m}
\end{aligned}
\)
where \(\Delta \mathrm{x}\) is the uncertainty in position and \(\Delta p_{x}\) (or \(\Delta \mathrm{v}_{\mathrm{x}}\) ) is the uncertainty in momentum (or velocity) of the particle. If the position of the electron is known with high degree of accuracy ( \(\Delta \mathrm{x}\) is small), then the velocity of the electron will be uncertain \(\left[\Delta\left(\mathrm{v}_{\mathrm{x}}\right)\right.\) is large]. On the other hand, if the velocity of the electron is known precisely \(\left(\Delta\left(\mathrm{v}_{x}\right)\right.\) is small), then the position of the electron will be uncertain ( \(\Delta \mathrm{x}\) will be large). Thus, if we carry out some physical measurements on the electron’s position or velocity, the outcome will always depict a fuzzy or blur picture.

Significance of Uncertainty Principle

Heisenberg Uncertainty Principle rules out the existence of definite paths or trajectories of electrons and other similar particles. If we know where a body is at a particular instant and if we also know its velocity and the forces acting on it at that instant, we can tell where the body would be sometime later. We, therefore, conclude that theWe conclude that the position of an object and its velocity fix its trajectory. Since for a sub-atomic object such as an electron, it is not possible simultaneously to determine the position and velocity at any precision, it is not possible to talk of the trajectory of an electron.

The effect of Heisenberg Uncertainty Principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects. This can be seen from the following examples.

If uncertainty principle is applied to an object of mass, say about a milligram \(\left(10^{-6} \mathrm{~kg}\right)\), then
\(
\begin{aligned}
\Delta \mathrm{v} \cdot \Delta \mathrm{x} &=\frac{h}{4 \pi \cdot m} \\
&=\frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times 3.1416 \times 10^{-6} \mathrm{~kg}} \approx 10^{-28} \mathrm{~m}^{-2} \mathrm{~s}^{-1}
\end{aligned}
\)

The value of \(\Delta \mathrm{v} \Delta \mathrm{x}\) obtained is extremely small and is insignificant. Therefore, one may say that in dealing with milligram-sized or heavier objects, the associated uncertainties are hardly of any real consequence.

In the case of a microscopic object like an electron on the other hand. \(\Delta \mathrm{v} . \Delta \mathrm{x}\) obtained is much larger and such uncertainties are of real consequence. For example, for an electron whose mass is \(9.11 \times 10^{-31} \mathrm{~kg}\)., according to Heisenberg uncertainty principle
\(
\begin{aligned}
\Delta \mathrm{v} \cdot \Delta \mathrm{x} &=\frac{h}{4 \pi m} \\
&=\frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times 3.1416 \times 9.11 \times 10^{-31} \mathrm{~kg}} \\
&=10^{-4} \mathrm{~m}^{-2} \mathrm{~s}^{-1}
\end{aligned}
\)

It, therefore, means that if one tries to find the exact location of the electron, say to an uncertainty of only \(10^{-8} \mathrm{~m}\), then the uncertainty \(\Delta \mathrm{v}\) in velocity would be
\(
\frac{10^{-4} \mathrm{~m}^{2} \mathrm{~s}^{-1}}{10^{-8} \mathrm{~m}} \approx 10^{4} \mathrm{~ms}^{-1}
\)
which is so large that the classical picture of electrons moving in Bohr’s orbits (fixed) cannot hold good. It, therefore, means that the precise statements of the position and momentum of electrons have to be replaced by the statements of probability, that the electron has at a given position and momentum. This is what happens in the quantum mechanical model of atom.

Problem 2.15
A microscope using suitable photons is employed to locate an electron in an atom within a distance of \(0.1 Å\). What is the uncertainty involved in the measurement of its velocity?

Solution:

\(\Delta \mathrm{x} \Delta p=\frac{h}{4 \pi}\) or \(\Delta \mathrm{x} m \Delta \mathrm{v}=\frac{h}{4 \pi}\)
\(
\begin{aligned}
&\Delta \mathrm{v}=\frac{h}{4 \pi \Delta \mathrm{x} m} \\
&\Delta \mathrm{v}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times 3.14 \times 0.1 \times 10^{-10} \mathrm{~m} \times 9.11 \times 10^{-31} \mathrm{~kg}} \\
&=0.579 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1}\left(1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right) \\
&=5.79 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}
\)

Problem 2.16
A golf ball has a mass of \(40 g\), and a speed of \(45 \mathrm{~m} / \mathrm{s}\). If the speed can be measured within accuracy of \(2 \%\), calculate the uncertainty in the position.

Solution:
The uncertainty in the speed is \(2 \%\), 1.e.,
\(45 \frac{2}{100}=0.9 \mathrm{~m} \mathrm{~s}^{-1}\)
Using the equation (2.22)
\(
\begin{aligned}
&\Delta \mathrm{x}=\frac{h}{4 \pi m \Delta \mathrm{v}} \\
&=\frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times 3.14 \times 40 g \times 10^{-3} \mathrm{~kg} \mathrm{~g}^{-1}\left(0.9 \mathrm{~ms}^{-1}\right)} \\
&=1.46 \times 10^{-33} \mathrm{~m}
\end{aligned}
\)
This is nearly \(\sim 10^{18}\) times smaller than the diameter of a typical atomic nucleus. As mentioned earlier for large particles, the uncertainty principle sets no meaningful limit to the precision of measurements.

Reasons for the Failure of the Bohr Model

In Bohr model, an electron is regarded as a charged particle moving in well defined circular orbits about the nucleus. The wave character of the electron is not considered in Bohr model. Further, an orbit is a clearly defined path and this path can completely be defined only if both the position and the velocity of the electron are known exactly at the same time. This is not possible according to the Heisenberg uncertainty principle. Bohr model of the hydrogen atom, therefore, not only ignores dual behaviour of matter but also contradicts Heisenberg uncertainty principle. In fact an insight into the structure of the atom was needed which could account for wave-particle duality of matter and be consistent with Heisenberg uncertainty principle. This came with the advent of quantum mechanics.