Quantum Physics

Quantum physics

The quantum theory of radiation:

The quantum theory of radiation was first proposed by plank in 1901 to explain the black body radiation. According, to this energy from the body is emitted in separate packets of energy each packet is called a quantum of energy. Each quantum carries a definite amount of energy called photon. Therefore, the energy carried by each photon is given by:

E = hf …..(i)

Where, ‘f’ is the frequency of radiation and ‘h’ is a constant called plank’s constant whose value is 6.62 * 10-34J/s.

This is the quantum theory of radiation. Therefore, from this relation (i) we know what the photons or quanta of high frequency has a large amount of energy while those of low frequency has less amount of energy.

Photoelectric effect:

The election of electrons from metallic surface when light is incident on it is known as photoelectric effect.

Einstein’s photoelectric equation:

According to Einstein’s (in 1905) quantum theory of radiation, light is a particle called quantum and the energy carried by each quantum s called photon. The rest mass energy of photon is zero. THe energy of photon having frequency ‘f’ is given by:

E = hf = hcλ$\frac{\mathrm{h}\mathrm{c}}{\lambda }$

Where, h = Plank’s constant

C = speed of light.

If light energy (photon) falls on any surface, it is used up in two ways by the surface.

(i) First part of energy called minimum energy is used to excite the electron in the atom and brings to the surface. This energy is called threshold energy; denoted by θ and given as:

Or, θ = f=hfo = hcλo$\frac{\mathrm{h}\mathrm{c}}{{\lambda }_{\mathrm{o}}}$

Where, f = threshold frequency of light.

(ii) The remaining part of light energy provides the K.E. of the emitted photoelectrons.

Ie. K.Emax = 12$\frac{1}{2}$mv2max

Where, m = mass of emitted photoelectron and,

Vmax = Maximum speed of emitted electron.

Therefore, we can write.

E = θ + K.E.max

Or, hf = hf1 + 12$\frac{1}{2}$mv2max

Or, h (f – fo) = 12$\frac{1}{2}$mv2max

This equation (i) is called Einstein’s photoelectric equation.

Milikan experiment to determine Plank’s constant:

Introduction:

An experiment set – up by Milikan to determine the value of Plank’s constant and hence to verify the Einstein’s photoelectric effect (equation) is called Milikan’s photoelectric experiment.

Experimental setup: The experimental set up for this experiment is shown as below.

Description:

It consists of a glass tube at the center which a rotating wheel W is present in which alkali metal (like Li, Na , K etc.) are coated. These alkali metal acts as cathode and cup shaped anode is present. These electrodes (anode and cathode) are connected through a variable battery and an electrometes(ammeter). A knife is used to scratch and remove the metal oxide from cathode (if formed). There is a window through which light passes the tube.

Working:

Light is allowed to fall of cathode. Due to the photoelectrons are emitted and move towards anode (being the potential). This is observed as photon current in electrometer ‘E’. Then the negative potential in begin to increases at anode ‘A’, this results the decrease in photocurrent for a particular value of – ve potential is called stopping potential denoted by Vs.

Therefore, we can write,

eVs = 12$\frac{1}{2}$mv2max = K.E.max…(i)

From einstein’s photoelectric equation, we write,

E = θ + K.E.max

Or, hf = hfo + eVs

Or, Vs = he$\frac{\mathrm{h}}{\mathrm{e}}$f - he$\frac{\mathrm{h}}{\mathrm{e}}$fo …(ii)

This, equation (ii) is a straight line of the form y = mk + c, where slope m = he$\frac{\mathrm{h}}{\mathrm{e}}$ and intercept c = −hfoe$-\frac{\mathrm{h}{\mathrm{f}}_{\mathrm{o}}}{\mathrm{e}}$ as shown in figure,

Now, perform the experiment with different frequency of light and corresponding stopping potential are observed. If the plots of Vs versus f is also of the form y = mx + c, then Einstein’s photoelectric equation is said to be verified.

Let, the slope of experimental result from graph be:

m = ba$\frac{\mathrm{b}}{\mathrm{a}}$ = changeinstoppingpotentialchangeinfrquency$\frac{\mathrm{y}}{}$

From equation (ii), we have,

Slope = he$\frac{\mathrm{h}}{\mathrm{e}}$

Or, ba=he$\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{h}}{\mathrm{e}}$

Or, h = ba$\frac{\mathrm{b}}{\mathrm{a}}$ * e.

Where, ‘e’ is the electronic charge.

Thus, knowing the value of a,b and e in above expression the value of Plank’s constant can be determined.

It’s value was found to be 6.62 * 10-34J/s.

Frank and Hertz experiment:

Introduction:

As experiment conducted by Frank and Hertz to show that the energy states of atom are quantized is known as Fran Hertz experiment.

Experimental set up:

The experimental set – up for this experiment is as shown below:

Description:

It consists of a glass tube (T) enclosing a gas vapor e.g mercury vapor. The glass tube also encloses three electrodes the cathode (c), the grid (G) and the plate (P). The cathode C is heated by filament F and the potential divides in connected across the cathode and the grid G in such a way that the grid is at + ve potential with respect to the cathode. The plate P is maintained at a small + ve potential with respect to the grid by connecting a small battery. The small – ve potential of P retard the electron.

Working:

When the power is switched on filament heat the cathode which then emits the electrons, the electrons are accelerated by the accelerating potential Va between the cathode C and the grid G. During acceleration an electron gains energy eVa and collides with the atom of mercury vapor. When Va is increased, more electron can relax the plate P overcoming the retarding potential (Vr). So, if Va is increased from zero, the plate current (Ip) rises, becomes maximum and falls suddenly to a minimum value when Va reaches a certain value Val – the excitation potential of the mercury atom.

So, frank and Hertz observed that for the mercury vapor in the tube, the current Ip falls suddenly for value of Va = 4.9V, 9.8V, 14.7V etc. This implies that the 1st excitation potential for mercury is 4.9ev from the observed fact that the plate current (Ip) falls suddenly at every internal of 4.9V of accelerating potential for mercury, they reached to the conclusion that the energy state in atom are quantized.

Bohr’s postulates:

The electron revolving round the nucleus only in certain definite circular orbits without radiation energy the possible orbit called the stationary of the atom.

The allowed states are those for which the orbital angular momentum of the electron mvr is equal to an integral multiple of h/2π

.i.e. mvrn =nh/2π (where r is the radius of the possible orbit m is the mass of electron revolving in that orbit, v be the velocity of the electron……..

Radiation of the energy hυ is emitted only when the electron jump from one stationary (ES2) sate of energy to another stationary state (ES2)

i.e. hυ= ES2- ES1

Let rn be the radius of nth orbit in which the electron revolved round the nucleus as shown in figure then,Then the coulomb force of attraction F between the charges provided the centripetal force mv2/rn to move the electron in the circular motion.

(where k=1/4π ε0 and z=1 for hydrogen atom)

Or, ke2rn$\frac{\mathrm{k}{\mathrm{e}}^2}{{\mathrm{r}}_{\mathrm{n}}^{}}$= mv2……1

On putting the value of v from the second postulate of Bohr’s we get

rn =n2h24mkπ2e2$\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4\mathrm{m}\mathrm{k}{\pi }^2{\mathrm{e}}^2}$ ……..2

on$\mathrm{o}\mathrm{n}$ Putting the value of k=1/4π ε0 in equation 2 we get

rn=n2h2ε0mπe2$\frac{{\mathrm{n}}^2{\mathrm{h}}^2{\epsilon }_0}{\mathrm{m}\pi {\mathrm{e}}^2}$ ……….4

This is the required expression for the radius of nthorbit.

The$\mathrm{T}\mathrm{h}\mathrm{e}$

Expresion of the radius of in the third orbit we can obtain on putting (n=3) in equation 4 we get

=32h2ε0mπe2$\frac{3^2{\mathrm{h}}^2{\epsilon }_0}{\mathrm{m}\pi {\mathrm{e}}^2}$=rn${\mathrm{r}}_{\mathrm{n}}$=9h2ε0mπe2$\frac{9{\mathrm{h}}^2{\epsilon }_0}{\mathrm{m}\pi {\mathrm{e}}^2}$

thistheexpressionofthe$\mathrm{t}\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{e}\mathrm{x}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{o}\mathrm{f}\mathrm{t}\mathrm{h}\mathrm{e}$Raduis for n=3

And the velocity of the electron in nth orbit is vn=nh/(2π mrn) by Bohr’s postulate

On putting the value of rn in this expression we get

 Vn=e22ε0nh$\frac{{\mathrm{e}}^2}{2{\epsilon }_0\mathrm{n}\mathrm{h}}$…………………5

Expression of energy of the electron in nthorbital

In each of the possible orbits, the electron will have definite energy given by the sum of potential energy and kinetic energy

Then the total energy of the electron in nth orbit (En)=K.E of electron + electrostatic potential energy

En= mv2/2 +-eV

En=e22x4πε0r−e24πε0r$=\frac{{\mathrm{e}}^2}{2\mathrm{x}4\pi {\epsilon }_0\mathrm{r}}-\frac{{\mathrm{e}}^2}{4\pi {\epsilon }_0\mathrm{r}}$ where z=1for hydrogen

En=−e28πε0r$-\frac{{\mathrm{e}}^2}{8\pi {\epsilon }_0\mathrm{r}}$……..6

On putting the value r or rn both are same in equation 6 we have

En=me48ε02n2h2$\frac{\mathrm{m}{\mathrm{e}}^4}{8{{\epsilon }_0}^2{\mathrm{n}}^2{\mathrm{h}}^2}$ ……………7

This is the energy of the electron in the hydrogen atom in the nth orbit.

Origin of spectral series of hydrogen atom:

The emitted light radiation when electron jumps from higher state to the lower state are called spectral linear.

A group of spectral lines are said to form spectral series if electrons jump from different excited states to a fixed lower state. The various spectral series of H are:

a. Lyman series: The spectral series formed when electrons jump from different higher states n2 = 2, 3, 4, 5,,,,as to a fixed lower state n1 = 1 i.e. ground state in called Lyman series. The wave length of this series for H – atom is given as:

or, [1λ=Ry(112−1n22),n2=2,3,4,5….]$\left[\frac{1}{\lambda }={\mathrm{R}}_{\mathrm{y}}\left(\frac{1}{1^2}-\frac{1}{{\mathrm{n}}_2^2}\right),{\mathrm{n}}_2=2,3,4,5\dots .\right]$

b. Balmier series: The spectral lines of this series correspond to the transition of an electron from some higher energy state to an orbit having n = 2. The wavelength of this series for H – atom is given by:

or, [1λ=Ry(122−1n22),n2=3,4,5..]$\left[\frac{1}{\lambda }={\mathrm{R}}_{\mathrm{y}}\left(\frac{1}{2^2}-\frac{1}{{\mathrm{n}}_2^2}\right),{\mathrm{n}}_2=3,4,\mathrm{5..}\right]$

The spectral line of this series corresponds to visible region.

1. Paschen series: The spectral lines of this series correspond to the transition of an electron from some higher energy state to an orbit n = 3. Therefore for Paschen series n1 = 3, n2 = 4,5,6….. The wavelength of this series for H – atom is given by:

Or, 1λ$\frac{1}{\lambda }$ = Ry(132−1n22)$\left(\frac{1}{3^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$

Paschen series lies in the infrared region of the spectral and it’s invisible.

d. Brackett series: The spectral lines of this series corresponds to the transition of an electron from a higher energy stare to the orbit n = 4.

Therefore, this series n1 = 4 and n2 = 5,6,7….The wavelength of this series for H – atom is given by:

Or, 1λ=Ry(142−1n22)$\frac{1}{\lambda }={\mathrm{R}}_{\mathrm{y}}\left(\frac{1}{4^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$

This series also lies in the infrared region of the spectrum.

e. P – fund series: The spectral lines of this series correspond to the transition of electron from a higher energy state to the orbit having n = 5.

Therefore, for this series, n1 = 5 and n2 = 6, 7, 8….The wavelength of this series for H – atom is given as:

Or, 1λ=Ry(152−1n22)$\frac{1}{\lambda }={\mathrm{R}}_{\mathrm{y}}\left(\frac{1}{5^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$

The END

Ry(152−1n22)