Atomic Structure 1.1

Video Available Study Notes PDF Resource

Video Lecture

Study Notes

1.1 Introduction

The atomic and molecular structure forms the backbone of chemistry, providing insight into how atoms bond and interact to form molecules.

1.2 Electromagnetic Radiation

Electromagnetic radiation is energy transmitted in the form of waves or particles across space or through a medium. It encompasses a broad range of wavelengths, from radio waves to gamma rays, but the most important in chemistry are visible light, ultraviolet (UV), and infrared (IR) radiation.

The energy \(E\) of electromagnetic radiation is given by Planck's relation:

\[E = h\nu = \frac{hc}{\lambda}\] where: \(h = 6.626\times 10^{-34}\ \mathrm{J\cdot s}\) (Planck’s constant) \(\nu\) is the frequency \(\lambda\) is the wavelength \(c = 3.0\times 10^{8}\ \mathrm{m/s}\) (speed of light)

Failure of Classical Mechanics

Till the end of the 19th century, Classical or Newtonian Mechanics was considered to be the only right and undisputed theoretical science. But soon, some new experimental phenomena were observed that could not be explained by Classical Mechanics. This includes:

  • Black Body Radiation
  • Photoelectric Effect

Classical mechanics gives the laws of motion of macroscopic objects, whereas quantum mechanics gives the laws of motion of microscopic objects.

Photoelectric Effect

The ejection of electrons from the surface of a metal by radiation is called the photoelectric effect.

Classically, electron ejection from the metal’s surface was thought to occur because of the oscillation of electrons with light, eventually breaking away from the surface with a kinetic energy dependent on the intensity of the incident radiation.

However, experimentally it was found that the kinetic energy of ejected electrons was independent of radiation intensity.

In 1905, Albert Einstein explained the photoelectric effect using Planck’s relation between energy and frequency. He proposed that light of frequency \(\nu\) consists of particles (photons) with energy \(h\nu\).

\[\mathrm{K.E.} = h\nu - W = h\nu - h\nu_0\]

This explained the observed dependence of electron energy on light frequency, giving light a particle character.

Key Conclusions

  1. Energy is quantized — transferable only in discrete packets called quanta.
  2. Light exhibits particle-like behavior.

Wave-Particle Dual Behavior

In 1924, Louis de Broglie proposed that if radiation has dual behavior, matter should too. He suggested particles have wave-like properties with wavelength:

\[\lambda = \frac{h}{p}, \quad p = mv\]

For macroscopic objects, the wavelengths are too small to observe. For example, a \(100\ \mathrm{kg}\) person walking at \(5\ \mathrm{m/s}\) has \(\lambda \approx 1.32\times 10^{-36}\ \mathrm{m}\).

Wave-like behavior of matter was confirmed experimentally by Davisson and Germer, who observed electron diffraction off a crystal lattice.

Heisenberg Uncertainty Principle

In 1926, inspired by wave mechanics, Heisenberg proposed the uncertainty principle:

\[\Delta x \cdot \Delta p \ge \frac{h}{4\pi}\]

This states that the position and momentum of a particle cannot be simultaneously determined with arbitrary precision.

Shorter wavelength light improves position measurement but increases momentum uncertainty, and vice versa.

Summary

  • Quantum mechanics governs microscopic matter.
  • Quantum chemistry explains structure, bonding, and reactivity.
  • Quantization: physical quantities can take only discrete values.
  • Quantum mechanics explains phenomena (e.g., blackbody radiation, photoelectric effect) that classical mechanics cannot.

Some Solved Important Questions

Q1. Calculate the de Broglie wavelength of alpha particles associated with an energy of \(100\ \mathrm{MeV}\).

\((1\ \mathrm{MeV} = 1.602\times 10^{-6}\ \mathrm{erg},\quad h = 6.626\times 10^{-27}\ \mathrm{erg\cdot s})\)

\(E = h\nu = \frac{hc}{\lambda} \quad\Rightarrow\quad \lambda = \frac{hc}{E}\) \(\lambda = \frac{(6.626\times 10^{-27})(3\times 10^{8})}{100\times (1.602\times 10^{-6})} \approx 1.24\times 10^{-14}\ \mathrm{m}\)

Q2. Calculate the energy per photon for radiation of wavelength \(650\ \mathrm{pm}\).

\(E = \frac{hc}{\lambda} = \frac{(6.626\times 10^{-34})(3\times 10^{8})}{650\times 10^{-12}} \approx 3.058\times 10^{-16}\ \mathrm{J}\)

Q3. Calculate the momentum of a particle with de Broglie wavelength \(1 \text{Angstrom}\).

\(h = 6.6\times 10^{-34}\ \mathrm{kg\cdot m^2/s}\)

\(p = \frac{h}{\lambda} = \frac{6.6\times 10^{-34}}{1\times 10^{-10}} = 6.6\times 10^{-24}\ \mathrm{kg\cdot m/s}\)

Q4. A moving electron has \(4.55\times 10^{-25}\ \mathrm{J}\) kinetic energy. Calculate its wavelength.

\(m = 9.1\times 10^{-31}\ \mathrm{kg},\quad h = 6.6\times 10^{-34}\ \mathrm{kg\cdot m^2/s}\)

\(v = \sqrt{\frac{2(4.55\times 10^{-25})}{9.1\times 10^{-31}}} \approx 10^{3}\ \mathrm{m/s}\) \(\lambda = \frac{6.6\times 10^{-34}}{(9.1\times 10^{-31})(10^{3})} \approx 7.25\times 10^{-7}\ \mathrm{m}\)

Q5. How many photons of wavelength \(4000\ \text{\AA}\) are needed to provide \(1.00\ \mathrm{J}\) of energy?

\(E_{\text{photon}} = \frac{hc}{\lambda} = \frac{(6.626\times 10^{-34})(3\times 10^{8})}{4000\times 10^{-10}} \approx 4.97\times 10^{-19}\ \mathrm{J}\) \(\text{No. of photons} = \frac{1.00}{4.97\times 10^{-19}} \approx 2.01\times 10^{18}\)

Q6. Electromagnetic radiation of wavelength \(242\ \mathrm{nm}\) ionizes sodium. Find ionization energy in \(\mathrm{kJ/mol}\).

\(E_{\text{atom}} = \frac{(6.626\times 10^{-34})(3\times 10^{8})}{242\times 10^{-9}} \approx 8.21\times 10^{-19}\ \mathrm{J}\) \(E_{\text{mol}} = (8.21\times 10^{-19})(6.022\times 10^{23}) \approx 4.94\times 10^{5}\ \mathrm{J/mol} \approx 494\ \mathrm{kJ/mol}\)

Q7. Calculate kinetic energy of an electron with wavelength \(4.8\ \mathrm{pm}\).

\(v = \frac{h}{m\lambda} \approx \frac{6.626\times 10^{-34}}{(9.1\times 10^{-31})(4.8\times 10^{-12})} \approx 1.516\times 10^{8}\ \mathrm{m/s}\) \(KE = \frac{1}{2} m v^2 \approx 1.047\times 10^{-14}\ \mathrm{J}\)

Q8. Electron speed \(600\ \mathrm{m/s}\) measured with \(0.005\%\) accuracy. Find minimum error in position.

\(\Delta v = 0.00005\times 600 = 0.03\ \mathrm{m/s}\) \(\Delta x \ge \frac{h}{4\pi m \Delta v} \approx 1.93\times 10^{-3}\ \mathrm{m}\)

Q9. Electron mass \(9.11\times 10^{-31}\ \mathrm{kg}\), position uncertainty \(10\ \mathrm{pm}\). Find uncertainty in velocity.

\(\Delta v \ge \frac{h}{4\pi m \Delta x} \approx 5.765\times 10^{6}\ \mathrm{m/s}\)

Q10. What is the designation of orbital with \(n=4,\ l=3\)?

\(4f\ \text{orbital}\)

Q11. Which quantum number specifies the shape of an orbital?

Azimuthal quantum number \(l\).

Q12. Why do we not observe wave properties of large objects?

Their wavelengths are too small to be observed.

PDF Resource

PDF Resource

Navigation