Physics > Basic Modern Physics > 3.0 Wave Particle Duality
Basic Modern Physics
1.0 Photon theory of light
2.0 Characteristics of photon
3.0 Wave Particle Duality
4.0 Emission of electrons
5.0 Photoelectric Effect
5.1 Laws of Photoelectric emission
5.2 Photoelectric equation
5.3 Photoelectric Current
5.4 Stopping potential
5.5 Graph between $K{E_{max}}$ and frequency
6.0 Radiation Pressure And Force
7.0 Photon Density
8.0 Force exerted by a light beam on a surface
9.0 Early Atomic Structures
10.0 Bohr Model of The Hydrogen Atom
10.1 Radius of Orbit
10.2 Velocity of electron in the $n^th$ orbit
10.3 Orbital frequency of electron
11.0 Energy of electron in the $n^{th}$ orbit
12.0 Basic Definitions
13.0 Atomic Excitation
3.2 Application of de Broglie Hypothesis
5.2 Photoelectric equation
5.3 Photoelectric Current
5.4 Stopping potential
5.5 Graph between $K{E_{max}}$ and frequency
10.2 Velocity of electron in the $n^th$ orbit
10.3 Orbital frequency of electron
- Quantization of orbit: If we consider a standing electron wave, then to maintain a standing wave over the circumference of a circular orbit, the wavelength must be an integral fraction of that circumference.$$\begin{equation} \begin{aligned} 2\pi r = n\lambda = \frac{{nh}}{p} = \frac{{nh}}{{mv}} \\ \Rightarrow mvr = \frac{{nh}}{{2\pi }} \\\end{aligned} \end{equation} $$
- Electron Microscope
Question 1. An electron and a photon have the same de Broglie wavelength. Which has higher kinetic energy?
Solution: Let $\lambda $ be the de Broglie wavelength of the electron and the photon. If $m$ and $v$ are the mass and velocity of the electron, the de Broglie wavelength of the electron $$\begin{equation} \begin{aligned} \lambda = \frac{h}{{mv}} \\ \\\end{aligned} \end{equation} $$
The photon has zero rest mass. Therefore energy of the photon is totally kinetic in nature. Since the wavelength of the photon is same as that of the electron, the kinetic energy of the photon having wavelength $\lambda $, $$\begin{equation} \begin{aligned} \quad \quad {E_1} = \frac{{hc}}{\lambda } = \frac{h}{{\left( {h/mv} \right)}} \\ or\quad {E_1} = mvc \\\end{aligned} \end{equation} $$
Now, the kinetic energy of the electron, $$\begin{equation} \begin{aligned} {E_2} = \frac{1}{2}m{v^2} = mv \times \frac{v}{2} \\ \\\end{aligned} \end{equation} $$
Since $c > \left( {v/2} \right)\;\;\left( {{\text{as }}c > v} \right)$, from the results (i) and (ii), it follows that $${E_1} > {E_2}$$
i.e., kinetic energy of the photon is greater than that of the electron. As it moves with the speed $c$, it is faster than electron.
Question 2. A photon and a electron have the same de Broglie wavelength. Which has greater total energy? Explain.
Solution: Let $\lambda $ be the de Broglie wavelength of the electron and the photon. If $m$ and $v$ are the mass and velocity of the electron, the de Broglie wavelength of the electron $$\begin{equation} \begin{aligned} \lambda = \frac{h}{{mv}} \\ \\\end{aligned} \end{equation} $$
Energy of photon having wavelength $\lambda $,
$$\begin{equation} \begin{aligned} {E_1} = \frac{{hc}}{\lambda } \\ \\\end{aligned} \end{equation} $$
Since, wavelength of the photon is same as that of the electron,
$$\begin{equation} \begin{aligned} {E_1} = hc \times \frac{{mv}}{h} = mvc \\ \\\end{aligned} \end{equation} $$
According to Einstein's mass-energy equivalence relation, the energy of the electron (mass $m$)
$${E_1} = m{c^2} = mc \times c$$
Since ,from Equation (i) and (ii), it follows that
$${E_1} > {E_2}$$
i.e, the total energy of electron is greater than that of the photon.
Note: In the previous problem the kinetic energy of photon and electron is compared while in this problem the total energy is compared.
Question 3. Find the ratio of de Broglie wavelength of molecules of hydrogen and helium which are at temperatures ${27^0}{\text{C}}$ and ${127^0}{\text{C}}$, respectively.
Solution: De Broglie wavelength is given by $\lambda = h/mv$.
Root mean square velocity of a gas particle at the given temperature ($T$) is given as,
$$\begin{equation} \begin{aligned} \frac{1}{2}m{v^2} = \frac{3}{2}kT \\ v = \sqrt {\frac{{3kT}}{m}} \\\end{aligned} \end{equation} $$
where $k$= Boltzmann's constant , $m$= mass of the gas particle and $T$= temperature of the gas in K.
$$\begin{equation} \begin{aligned} mv = \sqrt {3mkT} \\ \lambda = \frac{h}{{mv}} = \frac{h}{{\sqrt {3mkT} }} \\ \frac{{{\lambda _H}}}{{{\lambda _{He}}}} = \sqrt {\frac{{{m_{He}}{T_{He}}}}{{{m_H}{T_H}}}} = \sqrt {\frac{{4\left( {273 + 127} \right)}}{{4\left( {273 + 27} \right)}}} = \sqrt {\frac{8}{3}} \\\end{aligned} \end{equation} $$
