Nuclear, Quantum physics

Nucleus:

Nucleus was discovered by the Rutherford and his co-workers from the α- particle scattering experiment.

The various properties of nucleus are:

a. Each nucleus has a charge. If Z is the atomic number and e is the charge of an electron, the positive charge on a nucleus is q = +Z.

b. Each nucleus has a mass. If A is the mass number M is the mass of neutron, the total mass of a nucleus is given by M = mass of proton + mass of neutron.

So, M = ZMp + (A – Z)Mn

c. Each nucleus has its own size. The radius of a nucleus of mass number A is given by:

r = roA1/3, where ro = 1.15 * 10-13cm.

d. Each nucleus possesses binding energy.

e. Each nucleus has spinning motion about its own axis in the atom due to its protons and neutrons spinning.

f. Due to spinning motion, the nuclei have magnetic moments.

The properties of nucleus are:

g. The mass of nucleus is equal to the sum of the masses of proton and neutron.

Since the size of the nucleus is very small its density is high. Nuclear density is given by

  ρ =nuclearmassnuclearvoulme$\frac{\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{v}\mathrm{o}\mathrm{u}\mathrm{l}\mathrm{m}\mathrm{e}}$ = AMn43πR3$\mathrm{A}\frac{{\mathrm{M}}_{\mathrm{n}}}{\frac{4}{3}\pi {\mathrm{R}}^3}$ where Mn is the mass of the nucleus A is the mass number R is the radius of the nucleus.

h. The electric quadruple moment of the nucleus describes the interaction of the nucleus with the electric field and deviation of the nucleus from the spherical symmetry.

Einstein mass energy relation:

A/c to the relativistic theory mass energy relation is given by Einstein is “mass can be converted into energy and energy can be converted into mass and viceversa” and he give a relation E=mc2

This equation is known as mass energy equation. Where E be the energy produce, m be the mass of the object or material and c be the velocity of the light

a/c to the relativsity theory of the einstein we have a relation

m=m01−v2c2√$\mathrm{m}=\frac{{\mathrm{m}}_0}{\sqrt{1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}$ …………..1

where$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}$m be the relativistic mass of the body, m0 be the rest mass of the body, v be the velocity of the body and c be the velocity of the light.

Diffriatating eq. 1 we get

dm= ………..2

From eq. 1 and 2 we have

or, dm={ }

or, dm(c2-v2)=mvdv…………..3

From Newton second law of motion we have

F=rate of change of momentum=d(mv)/dt

Or, F = mdv/dt +vdm/dt………….4

Now K.E for the small displace ment(dx) is d(K.E) = F.dx……………..5

From 4 and 5 we have

d(K.E)=

Or, d(K.E)=mv dv +v2dm…………………6 where dx/dt=v

From equation 3 and 6 we have

d(K.E)=c2dm………7

on integrating eq. 7 from 0 to K.E and m0 to m we get

K.E=c2(m-m0)

Now the total energy of the body = rest energy + Kinetic enrgy(K.E)

Or, E=m0c2+c2m-m0c2=mc2

This is the Einstein mass energy relation.

Binding energy:

The total mass of stable nucleus is always less than the sum of the masses of its constituent by an amount ∆m called mass defect. The energy equivalent of this masses difference is referred to as the binding energy B i.e.

B=∆m C2=[{Zmp+(A-Z)mn}-m]C2

Where Z= atomic number

A=mass number

C2= velocity of light, mn=mass of neutron, mp= mass of proton and m=mass of nucleus.

Or. The energy is equal to this must be given to the nucleus to break it into itsconstituent protons and neutrons.

The average binding energy per nucleus is defined as the binding energy of a nucleus dividing by number of nucleons it contains. Figure show the binding energy per nucleon as the function of A.

Fig:

The properties of nuclear force are:

They are short ranged force.

They are strong force in nature

They have saturation properties

They are attractive force.

Fission:

The process of braking up of the nucleus of a heavy atom into two, more or less equal fragment with the releasing the large amount of energy is called fission. 

Nuclear fusion:

In this process two or lighter nuclei combination together to form a single heavy nucleus. For example when four hydrogen nuclei are fused together, the helium nucleus is formed. The mass of the single nucleus formed is always less than the sum of the masses of the individual light nuclei. The difference in mass is converted into energy a/c to Einstein’s equation

Different between nuclear fission and nuclear fusion:

Nuclear Fission	Nuclear Fusion
It is the process in which heavy (parent) nucleus split into lighter. nuclei (daughter)	It is the process in which lighter nuclei (daughter) combine to form heavy nucleus (partent)
It occurs when thermal neutrons bombard to heavy nucleus.	It occurs by heating lighter atom at high temperature.
Energy per nucleon is less than that of fusion. But energy release is greater than fusion.	Energy per nucleon is greater than that of fission. But energy released is fusion is less than that of fission.
During this process many radioactive isotopes may be formed.	During this process no radioactive isotopes are formed.
It is single stage reaction	It is multistage reaction.

Nuclei:

Isotopes are the nuclei having the same atomic number ‘Z’ but different mass no. ‘A’.

Nuclear density =mass of nucleus/ volume of nucleus

The volume of the nucleus is directly proportional to the mass no. A. if the no. of the nucleus is doubled, volume is also double. As the both mass and volume of the nucleus is increasing by the same ratio on increasing the no. of nucleus, hence the entire nucleuses have nearly the same density.

Binding energy:

the total mass of stable nucleus is always less than the sum of the masses of its constituent by an amount ∆m called mass defect. The energy equivalent of this masses difference is referred to as the binding energy B i.e.

B=∆m C2=[{Zmp+(A-Z)mn}-m]C2

Where Z= atomic number

A=mass number

C2= velocity of light, mn=mass of neutron, mp= mass of proton and m=mass of neucleus.

Or. The energy is equal to this must be given to the nucleus to break it into its constituent protons and neutrons.

The average binding energy per nucleus is defined as the binding energy of a nucleus dividing by number of nucleons it contains. Figure show the binding energy per nucleon as the function of A.

Characteristic of the curve:

1. for the lower atomic no. A, binding energy per nucleon is very low and rises as A increase and reaches a plateau of about8MeV per nucleon above A=20

2. There is maximum of about 8.8 MeV per nucleon around A=50

3. Binding energy per nucleon is almost a constant between A=20 and A=140 with the average Binding energy 8.5 MeV per nucleon.

4. Beyond at A=60 the curve decrease slowly with increasing A and reaches 7.6 MeV per nucleon at A=238. This indicates that the large nuclei are held together less tightly than that in the middle of the periodic table.

Construction and working of G – M tube or counter:

The device which is used to count the number of atoms decreasing per minute from any substance is called G – M tube or counter. With the help of this tube, we can also determine the half – life period and decay constant.

Construction:

Working:

Suppose one α – particle is decayed from radioactive substances and enters into the cylindrical tube. With increase in potential of H.T.B., α – particle accelerates and follows collision with gas atoms inside the tube. With further increase in potential of H.T.B., avalanche effect of electrons are generated by collision which results flow of current through R. Due to this potential is dropped across R. This potential drop is called pulse which amplified by amplifier and counted. Thus, one pulse represents disintegration of one atom.

Thus, by counting the number of pulse, the number of disintegration of atoms from radioactive substance can be counted for proper working of device, the voltage should be:

Or, v = v1+v22$\frac{{\mathrm{v}}_1+{\mathrm{v}}_2}{2}$

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$\frac{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{p}\mathrm{p}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}{\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{i}\mathrm{n}\mathrm{f}\mathrm{r}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}}$

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.

Thrfore F = ke2r2n=mv2rn$\frac{\mathrm{k}{\mathrm{e}}^2}{{\mathrm{r}}_{\mathrm{n}}^2}=\frac{\mathrm{m}{\mathrm{v}}^2}{{\mathrm{r}}_{\mathrm{n}}}$  (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 nth orbital

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 principle of laser:

The laser light is a source that produces a beam of highly coherent and a very monochromatic light as a result of a co – operative emission from many atoms.

Principle of laser: Let is consider an assembly of atoms of some kind ‘hf’. If we somehow raise the atoms of the metastable level and let a light of frequency ‘f’ fall upon them, there will be more induced emission from the metastable level than that of induced absorption by the lower level. As a result an amplification of the original light is obtained. This is the principle of laser.

Ruby laser:

The ruby is a crystal of Aluminum oxide (Al2O3) mixed with it 0.05% of chromium oxide (Cr2O3).

Construction: A ruby laser consists of a cylindrical ruby rod whose ends are optically flat. One end is fully silvered and the other end is partially silvered. The ruby rod is enclosed in a glass tube which is surrounded by a xenon flash tube which acts as an optical pumping device in figure.

Working: The chromium atoms are excited from E1 state to E3 state by absorption of light of wavelength 350 nm produced by xenon flash tube. The excited chromium atoms then undergo non radioactive transition to E2 state which is metastable state having means time about 10-3 sec. When population inversion (N2> N1) is achieved, some of the atoms undergo spontaneous transition from E2 to E1, thus emitting photons of frequency f = E3 – E1/h. The photons moving not parallel to the ruby rod escapes out and those moving not parallel to the ruby rod escape out and those moving parallel to the rod gets deflected back by the silvered surface. The photons which are moving parallel to the ruby rod can cause the stimulated emission. As a result, a beam of coherent photons moving in phase and parallel to the ruby rod comes out of the partially silvered end of the rod.

Population inversion:

Let N1 and N2be the number of atoms lying in the ground state E1 and E2 respectively. Under ordinary condition of thermal equilibrium the number of atoms in the higher energy state is considerably smaller than that of the number of atoms in the lower energy state is N2< N1. In such a situation of a light of frequency f = E2−E1h$\frac{{\mathrm{E}}_2-{\mathrm{E}}_1}{\mathrm{h}}$ is incident on a large collection of such atoms, the atoms are excited due to absorption of photons and raise to the excited state E2 than that in the lower energy state E1 i.e. N2> N1. This phenomenon of having more number of atoms in the excited state than in the ground state is called population inversion.

The process by which population inversion is carried out is called optical pumping.

Optical pumping is a method of achieving population inversion. For this atom which have three energy states E1, E2 and E3 are taken when E3> E2> E1. E1 is a ground state, E2in metastable state and E3 is short lived state. The atoms in the E1 state are pumped into E3 stated by photon of energy hf = E3 – E1 by stimulated absorption. E3 is shortly lived state and E3 to E1 transition is prohibited. Since, N3> N2 cannot be achieved, so simulated emission results from E3 to E2 transition. Thus, the atoms in the atoms in the E3 state then go to the E2 state either by simultaneous emission or by non radioactive transition in which the energy E3 – E2 is converted into vibration energy of the atom forming the substance. Since E2 is metastable state, atoms can remain in this state for comparatively longer time. As a result, population inversion (i.e. N2> N1) takes place. Now, the atoms in E2 state are bombarded by a photon of energy hf = E2 – E1 to cause them to a stimulated emission of radiation of energy hf in the direction of incident photon. As a result a highly intense coherent and unidirectional beam of radiation comes. The beam is called laser beam.

Construction of He-Ne laser:

It consist of a long and narrow discharge tube about 8ocm long and 1cm in diameter, fill with the mixture of Heat a pressure of 1mm and Ne at a pressure 0.1mm of Hg, which forms the leaser medium. Two electrodes P and Q fitted to the discharge tube as shown in figure M and N are the two mirrors which form a resonant cavity. The mirror M is fully reflecting whereas N is partially reflecting so, allow the laser beam to pass out of it.

Working:

 As soon as the electric discharge passes through the mixture of He and Ne gases electron in the tube are accelerated. Theses collide with helium atom and excited them to higher energy level S2 and S3.as shown in figure. These levels of the helium atoms are metastable and the excited helium atom s remains in these for a long time before being de-excited. On the other hand, some of the excited energy state, say E4and E6 of neon correspond very approximately to the energy state S2 and S3.

Because of this when helium atoms in the energy state S2and S3 collide with neon atoms in the ground state E1, Ne atom absorb energy and are excited to the energy state E4and E6whereas He atom lose energy and are de-excited to the ground state S1. This process is continuously transfer more and more neon atom from the ground state to the excited energy sate E4 and E6

There are to more energy E3 and E5 slightly below the sates E4and E6 and another energy state E2 in Ne. Since the energy state E4and E6 are highly populated, there is a population inversion between the state E4and E6and the lower energy sate E3and E5. And the emission of the radiation are transition between E6and E5,E4 and E3, and E6and E2 respectively lead to emission of wavelength 3.539μm, 1.15μm and 6328A0. The first two lies on the infra red region and the last cross ponds to the red light from He-Ne laser in the visible region.

In an improved of the He-Ne laser, the resonator mirror, P and Q are used externally to the lase cavity because the mirror sealed inside the discharge tube are eroded by gas discharge and have to replaced.

Further for the minimization of the loss due to reflection at the ends of the discharge tube, Brewster windows W1 and W2 are used as shown in figure these are fitted at an angle called Brewster angle and given by tan⁡ɵB=μ

Where μ is the refractive index of the material of which the window are made.

Uses:

It is used in holography

It is used in scientific research.

It is used in medical treatment.

Heisenberg’s uncertainty principle:

In classical mechanics position co-ordinates, component of momentum, components of angular momentum are measured in an arbiter precision but in quantum mechanics particle is considered as wave and components of position and momentum are measured with highly improved precision as compared to classical mechanics.

The amplitude of the wave remaining almost zero in wide range of wave but specific value in narrow region of wave indicates high precision in position of the particle measured as momentum is conjugate quantity with position, high precision in position reflects worst precision in momentum. Hence, Heisenberg states that ‘it is impossible to determine position and momentum simultaneously with 100% accuracy.

These quantities are related as:

Δx. Δp ∼ h

Where Δx is uncertainty in position

Δp is uncertainty in momentum

Planks constant

This equation was further improved by Heisenberg which can be expressed as

                                            Δx.Δp≥h2π$\mathrm{\Delta }\mathrm{x}.\mathrm{\Delta }\mathrm{p}\ge \frac{\mathrm{h}}{2\pi }$

Uncertainty principle proposed by Heisenberg was further modified by Kennard and obtained the result as:

Δx.Δp≥h4π$\mathrm{\Delta }\mathrm{x}.\mathrm{\Delta }\mathrm{p}\ge \frac{\mathrm{h}}{4\pi }$

The most important thing is that uncertainty principle holds true for both microscopic and microscopic particle and it is independent with the method of measurement.

Electron cannot be constituents of nucleus:

Let electron lie inside the nucleus. Then its uncertainty in position is equal to diameter (size) of nucleus which is about 2 * 10-14m.

So, uncertainty in position of electron,

Or, Δ$\mathrm{\Delta }$x = 2 * 10-14m

Using, Heisenberg’s uncertainty principle the uncertainty in momentum of electron is given by:

Or, ΔP$\mathrm{\Delta }\mathrm{P}$ = h2πΔn$\frac{\mathrm{h}}{2\pi \mathrm{\Delta }\mathrm{n}}$

So, Uncertainty in speed of electron is:

Or, ΔV=ΔPme$\mathrm{\Delta }\mathrm{V}=\frac{\mathrm{\Delta }\mathrm{P}}{\mathrm{m}\mathrm{e}}$

= h2πΔn.me$\frac{\mathrm{h}}{2\pi \mathrm{\Delta }\mathrm{n}.{\mathrm{m}}_{\mathrm{e}}}$

= 6.62∗10−342π∗2∗10−14∗9.1∗10−31$\frac{6.62\ast {10}^{-34}}{2\pi \ast 2\ast {10}^{-14}\ast 9.1\ast {10}^{-31}}$

= …… ms-1 >>>> C

Where, C is the speed of light. Since, speed of electron cannot be greater than speed of light, electron cannot be constituents of nucleus.