Explore topic-wise InterviewSolutions in .

CONTENT:

Specific heat:

• It is defined as theamount of heat required to increase the temperature by 1°Kfor 1 kgof material.

⇒Q = mC∆t -----(1)

Where Q(J) = quantity of heat absorbed by a body, m(kg) = mass of the body, ∆t(°K) = Rise in temperature, and C = Specific heat

• S.I unitof specific heat ist isJ kg-1K-1.

Photon:

• According to Einstein's theory,the light propagates in the bundles of energy, each bundle is being called aphoton.
• ​Therest mass of the photon is zero. But its effective mass is given as,
• Photonexerts pressure on the surface.

• The energy of a photon is given as,

⇒ E = hν

Where, h = Planck's constant, and ν= frequency

CALCULATION:

Given (Specific heat) C= 4.2J/gº-C, (mass) m = 500 g, T1= 30° C,T2= 50° C, and(frequency) ν =2×109 Hz

• We know that the energy of one photon is given as (h = 6.6×10-34J-sec),

⇒ E = hν

⇒ E =6.6 × 10-34× 2 × 109

⇒ E = 13.2 × 10-25J -----(1)

• The change in temperature of the water is given as,

⇒ΔT =T2- T1

⇒ΔT = 50 - 30

⇒ΔT = 20°C

• The heat needed for 20°C temperature change for 500 gm of water is given as,

⇒ Q = mCΔT

⇒ Q = 500 × 4.2 × 20

⇒ Q = 42000 J -----(2)

• If N number of the photon is required to change this temperature, then,

⇒ NE = Q

\(⇒ N=\frac{42000}{13.2×10^{-25}}\)

⇒ N =3.2 × 1028

• Hence, option 2 is correct.

CONCEPT:

• de Broglie wavelength of electrons:Louis de Broglietheorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
  • The wavelength of material waves isalso known as thede Broglie wavelength.
• de Broglie wavelength of electronscan be calculated from Planks constanth divided by the momentum of the particle.

λ = h/p

whereλ is de Broglie wavelength,h is Plank's const, and p is the momentum.

• In terms of Energy,de Broglie wavelength of electrons:

\(λ=\frac{h}{\sqrt{2mE}}\)

whereλ is de Broglie wavelength,h is Plank's const, m is the mass of an electron, and E is the energy of the electron.​

Theequation ofenergyof photons is given by:

\(E=\frac{hc}{λ}\)

where E isenergy, h is the Planck constant, c is the speed of light in a vacuum and λ is the photon's wavelength.

EXPLANATION:

Given that:

\(\lambda_e ={ h\over\sqrt{( 2 mE}) }~and~\lambda_p = {hc \over E}\)

\(\lambda_e^2 ={ h^2\over{( 2 mE}) }\)

\(\lambda_e^2 ={ h^2\over{ 2 m({hc \over \lambda_p}}) }\)

\(\lambda_e^2 ={ h^2\over{ 2 m({hc }}) } \lambda_p\)

\(\lambda_e^2 \propto \lambda_p\)

Sothe correct answer isoption 3.

CONCEPT:

• Thework function istheminimum amount of energy required to releasetheelectronfrom a photoemissive surface and is given by

⇒ ϕ = hν0

Where h = Plancks constant,ν0= threshold frequency

• TheEnsitens photoelectricequationgivesthekinetic energy of a photoelectron, thekinetic energy ofaphotoelectron is the difference in energy betweentheincident photon and work functionof the material and is given by

⇒Kinetic energy(KE) = hν -ϕ

Whereν = frequency of incident light, h = Plancks constant,ϕ = Work function

EXPLANATION:

• The kinetic energy of photoelectron is given by

⇒KE = hν -ϕ

\(⇒ \frac{1}{2}mV^{2} = hν -\phi \)(For incident frequencyν )

• If the frequency is doubled then the above equation can be written as

\(⇒ \frac{1}{2}mV^{2} = 2hν -hν_{0}\)

Assumeν =ν0, the above equation can be written as

\(⇒ \frac{1}{2}mV^{2} = 2hν -hν = hν \)

⇒ mV2= 2 hν

\(⇒ V = \sqrt{\frac{2h\nu}{m}}\)

• Hence, option 2 is the answer

CONCEPT:

Photoelectric Effect:

• ​When the light of asufficiently small wavelength is incidenton the metal surface,electrons are ejected from the metal instantly. This phenomenon is called thephotoelectric effect.​
• Einstein’s equation of photoelectric equation:

⇒KEmax= hν-ϕo

Where h = Planck's constant,ν = frequency of incident radiation, ϕo= work function, and KE =maximum kinetic energy of electrons.

• The relationship between kinetic energy and stopping potential is

⇒KE = qV0

Where KE = Kinetic energy, q = charge, V0= Stopping potential

CALCULATION:

Given -KEmax= 0.5 eV , e = 1.6 × 10-19C

• Stopping potential only depends on the maximum KE of an electron ejected
• The stopping potential is given by

⇒KEmax= eV0

Substituting the given values in the above equation

⇒0.5eV = 1.6 × 10-19× V0

\(\Rightarrow V_{0} = \dfrac{0.5\times 1.6\times 10^{-19}}{1.6\times 10^{-19}} = 0.5V\)

• Hence, option 4 is the answer

CONCEPT:

• Photonis an elementary particle whichsupplies energy to the electronsin a metal surface andby absorbingthisenergytheelectron comes out ofthemetal surface
• The rest mass of the photon iszero
• Proton is a positively charged nucleon
• The energy of an elementary particle in terms of wavelength can be written as

\(⇒ E = \frac{hc}{λ}\)

Where h = Plancks constant andλ = Wavelength

• The wavelength of an electroncan be written as

\(λ =\frac{h}{\sqrt{2ME}}\)

Where h = Plancks constant, M = mass of the proton, E = kinetic energy

EXPLANATION :

• The wavelength of an electron can be written as

\(⇒ λ _{e}=\frac{h}{\sqrt{2M_{e}E}}\)----(1)

• The energy of an elementary photonin terms of wavelength can be written as

\(⇒ E = \frac{hc}{λ_{p}}\)------(2)

Squaring equation 1 and substituting the value E from equation 2 to equation 1 (Since both the particle has the same energy)

\(⇒ \lambda^{2}_{e} = \frac{h^{2}}{2M_{e}\frac{hC}{\lambda_{P}}}\)

\(\Rightarrow \lambda_{e}^{2} = \frac{h^{2}\times \lambda_{P}}{2M_e hC}\)

The above equation can be written as

\(\Rightarrow \lambda_{e}^{2} \propto \lambda_{P}\)

• Hence, option 4 is the answer

CONCEPT:

• Photoelectric Effect: Certain metals eject free electrons when a ray of light strikes them. This phenomenon is known as the Photoelectric effect.
• Einstein equation for the photoelectric effect is given as

K = hν -ϕ

ν is the frequency of light striking metal surface, h is Planck's constant,ϕ is work function.

• Work function (ϕ): The minimum energy required to emit electrons out of metal is known as the work function.
• Cutoff Frequency: The minimum frequency required bya ray of light to emit electrons is known as theCutoff Frequency.

ϕ = hν0

whereν0 is the cutoff frequencyfor a light ray to emit electron out of given metal.

Kinetic Energy: The energy possed by a body in motion is known as Kinetic Energy.

\(K = \frac{1}{2} mv^2 \)

m is mass of the body, v is the velocity of the body

CALCULATION:

Given

Cuttoff frequency = ν

Frequency of light ray striking = 2ν

The energy of each photon falling = 2hν

the Maximum kinetic energy K obtained from the metal havingcutoff frequency is v.

K = 2hν - hν (By Einestein Equation)

⇒ K = hν

⇒ \(K = \frac{1}{2} mv^2 = h\nu\)

⇒\(v^2 = \frac{ 2h\nu}{m}\)

⇒\(v =\sqrt{\frac{ 2h\nu}{m}} \)

CONCEPT:

• de Broglie wavelength of electrons:Louis de Broglietheorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
  • The wavelength of material waves isalso known as thede Broglie wavelength.
• de Broglie wavelength (λ)of electronscan be calculated from Planks constanth divided by the momentum of the particle.

λ = h/p

whereλ is de Broglie wavelength,h is Plank's const, and p is the momentum.

• In terms of Energy,de Broglie wavelength of electrons:

\(λ=\frac{h}{\sqrt{2mE}}\)

whereλ is de Broglie wavelength,h is Plank's const, m is the mass of an electron, and E is the energy of the electron.

CALCULATION:

Given that E2= 2E1

\(λ_1=\frac{h}{\sqrt{2mE_1}}\) .....(i)

\(λ_2=\frac{h}{\sqrt{2mE_2}}\) .....(ii)

\(\frac{λ_1}{λ_2}=\sqrt\frac{E_2}{E_1}=\sqrt\frac{2E_1}{E_1}\)

\(\frac{λ_2}{λ_1}=\sqrt\frac{1}{2}\)

\({λ_2}={λ_1}\frac{1}{\sqrt2}\)

So the correct answer isoption 2.

CONCEPT:

EXPLANATION:

• From the above, it is clear that themathematical form of Maxwell Boltzmann Law is

\(n_i=\frac{g_i}{e^{\alpha+\beta\epsilon_i}}\)

• Hence, option 3 is correct.

Concept:

Force exerted on the surface is given by the following formula

\(F = \left( {1 + r} \right)\frac{{IA}}{C}\)

where

r is the Reflected radiation

A is the Surface area

c is the speed of the sound

Calculation:

Given,

Energy density, I = 50 W/m2

Speed of the sound, c = 3 × 108 m/s

Surface area, A = 1 m2

Reflected radiation, r = 25% = 25/100 = 0.25

Force on the surface (25% is reflecting back and the remaining 75% has been absorbed from the surface)

Force exerted on the surface is given by,

\(F = \left( {1 + r} \right)\frac{{IA}}{C}\)

\(= \frac{{\left( {1 + 0.25} \right) \times 50 \times 1}}{{3 \times {{10}^8}}}\)

\(= 1.25 \times \frac{{50}}{{3 \times {{10}^8}}}\)

\(= \frac{{125}}{{100}} \times \frac{{50}}{{3 \times {{10}^8}}}\)

\(= \frac{{125}}{6} \times {10^{ - 8}}\)

≃ 20 × 10-8 N

The correct answer is option 2) i.e.Field emission

CONCEPT:

• Electron emission:Itis the phenomenon when electronsescape from a metal surface. There are three types of electron emission.
  1. Thermionic Emission: It is the process in which the electrons are ejected from a metal surface when the metal is heated to a sufficient temperature.
  2. Field Emission: It is the process in whichelectrons are pulled out of the metal surface due to a very strong electric field.
  3. Photoelectric effect:The photoelectric effect is a phenomenon whereelectrons are ejectedfrom ametal surfacewhen thelight of sufficient frequencyis incident on it.

EXPLANATION:

• ​Electrons emitted due to the electric field applied to a metal surface is a result of field emission. Therefore, the correct answer is field emission.

CONCEPT:

• Theminimum energyrequiredto remove an electron from the ground state to outside the atomis calledionization energy.

The energy of electrons in any orbit is given by:

\({\bf{Energy}}\;{\bf{in}}\;{\bf{nth}}\;{\bf{orbit}}{\rm{\;}}\left( {{E_{n\;}}} \right) = \; - 13.6\;\frac{{{Z^2}}}{{{n^2}}}\;eV\)

Where n = principal quantum number and Z = theatomic number

• This is the energy that is required to remove an electron from its orbit.

CALCULATION:

• The minimum energy that is given to the electron and leaves the atom, this energy is known as the ionization energy.
• So the correct answer isoption 3.

Correct option-4

Concept:

• Quantum mechanics:It is a branch of mechanics deals with the study ofmotion and interactionofsubatomic particleslike theelectron, proton, neutrons,etc. with the help of mathematical descriptions (linear algebra, vector space, differential equations, special integrations, etc.)
• InQuantum mechanicsthere are somewave quantitiesassociated withquantum particlesorsubatomic particles.
• The most popular symbol is used to represent the wave quantity associated with thequantum particles is\(\psi (r,t)\)whatwe call the wave function of the particle.
• The wave function\(\psi (r,t)\)represents thestateof a quantum particle in quantum mechanics at a given instant of time.
• The stateof a Quantum particle in quantum mechanics at any instant of time is described by two quantities, position and velocity.

Calculation:

In quantum mechanics, a quantum particle in a given system is treated as a wave-particle.

Actually, they are assumed as wave packets which is a linear combination of several waves that interfere constructively.

The energy associated with such wave-particle is given by-

\(E=h\vartheta \) ----(i)

Since\(\vartheta =\frac{ω }{2\pi }\) ----(ii)

where,

ω is the angular frequency

On substituting the given value of\(\vartheta\)from equation (ii) in equation (i), we get

\(E=h\frac{ω }{2\pi }=\hbar ω \) Here\(\hbar =\frac{h}{2\pi }\)

\(\therefore E=\hbar ω \)

Thus, option-4is the correct answer.

CONCEPT :

• De Broglie wavelength connects between the wavelength and momentum of the particle and is given by, and is given by

\(⇒ λ = \frac{h}{P} = \frac{h}{mV}\)----(1)

Where m = mass of the particle and V = Velocity, h = Plancks constant

Substituting the value\(P = \sqrt{2mE}\)in equation 1, it can be rewritten as

\(⇒ λ = \frac{h}{\sqrt{2mE}}\)

Where m = mass, E= kinetic energy,

EXPLANATION :

• The de Broglie wavelength is given by

\(⇒ \lambda =\frac{h}{P}\)

or\(λ=\frac{h}{\sqrt{2mE}}\)--------(1)

• The kinetic energy of particle accelerated through potential V is given by

⇒ E = q × V

Where q = Charge, V = potential difference

For an alpha particle q = 2e

⇒E = 2e × V

and M = 4mP

Substituting the given values in equation 1

\(⇒ λ=\frac{h}{\sqrt{4m_P × 2e× V}}\)

Substituting the given values mp= 1.67 × 10-27Kg, e = 1.6 × 10-19C, h = 6.626 × 10-34Joues/sec and solving we get

\(⇒ λ=\frac{0.101}{\sqrt{V}}\)

• Hence, option 2 is the answer

The correct answer is option 3) i.e.Photon has a fixed mass.

CONCEPT:

• Photon:Photon is the elementary unit that makes up light. It is the discrete amount ofelectromagnetic energy.
• Theproperties of a photonare:
  1. It has no mass.
  2. It does not have any electrical charge.
  3. It can becreated or destroyed only when radiation is emitted or absorbed.
  4. It is not affected by the magnetic field and electric field.
  5. A photon travels with a speed = 3 x 108m/s.

​EXPLANATION:

• Photon ismasslessand therefore, option 3) is the incorrect statement.

Concept:

Heisenberg’s Uncertainty Principle:

• W Heisenberga German physicist in 1927, stated the uncertainty principle which is the consequence of dual behavior of matter and radiation.
• It states thatit is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.

Explanation:

According toHeisenberg’s Uncertainty Principle

\({\rm{\Delta }}x \times {\rm{\Delta }}P\; \ge \frac{h}{{4\;\pi \;}}\)

Where, Δx = Uncertainty in position, ΔP = Uncertainty in momentum, h = Plank’s constant

The correct answer is Black.

• A solar cooker is painted black from the outside.

• The black colour is used on the outer side because the black colour is known as the perfect absorber.
• A solar cooker is a device that uses the energy from direct sunlight to heat, cook food materials.
• Sunlightis radiation generated by fluctuating electric and magnetic fields.
• When the photons of light waves interact with molecules of the substance the sunlight gets convertedto heat.
• The sunlightemitted by the sun possesses electromagnetic radiationin them.
• When thesunlight strikes, the energy causes the molecules of the matter to vibrate and the molecules get excited and jump to higher levels. This activity generates heat.
• A concave mirror is used in solar cookers.

A quantum is defined asthe minimum amount of any physical entityinvolved in an interaction. For example, a photon is a single quantum of light.

Photon:

1)A photon is a tiny light particle that comprises waves of electromagnetic radiation.

2)James Maxwellobserved photons are just electric fields traveling through space.

3)Photons haveno charge and no resting mass.

4)It travels at thespeed of light.

Theenergy of Photonis given by:

E = hν

Where,

h is Planck's constant,and

νis the frequency of the Electromagnetic wave associated with the photon

h = 6.626 070 15 x 10-34J Hz-1

Positrons:

• Positron, also calledpositive electron, positively charged subatomic particlehaving the same mass and magnitude of chargeas the electron and constitutingthe antiparticleof a negative electron.
• The first of the antiparticles to be detected, positrons were discovered byCarl David Anderson.

Phonons:

• The thermal vibrations resulting from the thermal energy of the substances are called phonons.
• They can't absorb or emit electromagnetic radiation.

Optical phonons:

• The quantum of the lattice vibration in opposite directions are called optical phonons.
• The optical phonons are observed for the crystals containing multiple atoms as a basis such as NaCl, GaAs, ZnO, etc.

PhantomLoading:

• It is the phenomenon in which the appliances consume electricity even when they turn off.
• The phantom loading is used for examining the current rating ability of the energy meter.
• The actual loading arrangement will waste a lot of power.
• The phantom loading consumes very less power as compared to real loading, and because of this reason, it is used for testing the meter.

Relation between Relativistic E and\(\vec{P}\)

1) E = K + moc2

2)\(E = \dfrac{m_oc^2}{\sqrt{1-\dfrac{r^2}{c^2}}}\)

3)\(P = \dfrac{m_o}{\sqrt{1-\dfrac{r^2}{c^2}}}\)

Derivation:

We begin by squaring equation (2) on both sides, i.e.

\(E^2 = \dfrac{m_o^2c^4}{1-r^2/c^2}\)

Next, we insert the term (V2 - V2) as shown:

\(E^2 = \dfrac{m_o^2 c^2 (V^2 - V^2 +c^2)}{1-r^2/c^2} \)

\(=\dfrac{m_o^2c^2V^2}{1-r^2/c^2} - \dfrac{m_o^2 c^2 V^2}{1-r^2 /c^2} + \dfrac{m_o^2 c^4}{1-r^2/c^2}\)

\(E^2=\underset{= \ p}{\underbrace{\left(\dfrac{m_oV}{\sqrt{1-r^2/c^2}}\right)^2}}c^2 + \dfrac{m_oc^2 c^4 - m^2 c^2 r^2}{1-r^2/c^2}\)

\(E^2 =p^2 c^2 + m_o^2 c^2 \left(\dfrac{c^2 - r^2}{1-r^2/c^2}\right)\)

\(E^2 = p^2 c^2 + m_o^2 c^2 \left(\dfrac{c^2 - V^2}{\dfrac{c^2 - V^2}{C^2}}\right)\)

\(E^2 = p^2 c^2 + m_o^2 c^4\)where E is the total energy of the particle

\(m_o^2 c^4 = E^2 - p^2 c^2\)← will be identical in all inertial reference frames.

Concept:

In one dimension, wave functions are often denoted by the symbol ψ(x,t).

They are functions of the coordinate x and the time t. But ψ(x,t) is not a real, but acomplex function, the Schroedinger equation does not have real, butcomplex solutions.

The wave function of a particle, at a particular time, contains all the information that anybody at that time can have about the particle.

But the wave function itself has no physical interpretation. It is not measurable. However, the square of the absolute value of the wave function has a physical interpretation.

In one dimension, we interpret |ψ(x,t)|2as a probability density, a probability per unit length of finding the particle at a time t at position x.

The probability of finding the particle at time t in an interval ∆x about the position x is proportional to |ψ(x,t)|2∆x.

This interpretation is possible because the square of the magnitude of a complex number is real.

For the probability interpretation to make sense, the wave function must satisfy certain conditions. We should be able to find the particle somewhere, we should only find it at one place at a particular instant, and the total probability of finding it anywhere should be one. This leads to the requirements listed below.

• The wave function must be single valued and continuous. The probability of finding the particle at time t in an interval ∆x must be some number between 0 and 1.
• We must be able to normalize the wave function. We must be able to choose an arbitrary multiplicative constant in such a way, so that if we sum up all possible values ∑|ψ(xi,t)|2∆xi we must obtain 1.

CONCEPT:

EXPLANATION:

• In the photoelectric experiment, light behaves as a particle.
• When light undergoes diffractionand refraction it behaves as a wave. Hence, diffraction and refraction indicate that light is a wave.Therefore, option 1 is correct.

Concept:

Heisenberg’s Uncertainty Principle:

• 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

\({\rm{\Delta }}x \times {\rm{\Delta }}{p_x} \ge \frac{h}{{4\pi }}\)

∆x is uncertaintyin position,∆px is uncertaintyin momentum at position x.​

Explanation:

Now, it is given thatuncertainty in momentum is zero.

∆px = 0

But, by the uncertainty principle

\({\rm{\Delta }}x \times {\rm{\Delta }}{p_x} \ge \frac{h}{{4\pi }}\)

\(\implies {0} \times {\rm{\Delta }}{x} \ge \frac{h}{{4\pi }}\)

\(\implies {\rm{\Delta }}{x} \ge \frac{h}{{4\pi \times 0 }} \)

Now anything divided by zero is infinity. So,∆x is infinity.

So, infinity is the correct answer.

Concept:

• Light is said to be made up of particles with zero rest mass known as a photon.
• Theenergy of a photonis given by the equation:

E = hν -- (1)

h is Planck's constant,ν is the frequency of light

The frequency of light can be expressed as the ratio of the speed of light c and its wavelength.

\(\nu = \frac{c}{λ}\)-- (2)

From (1) and (2) we can write

\(E = \frac{hc}{λ}\)-- (3)

• 1Ȧ = 10 -10 m

Calculation:

Given,

Planck constant h =6.6 × 10-34joule-second

wavelength λ = 3300 Ȧ

So, the energy will be

\(E = \frac{hc}{λ}\)

\(\implies E = \frac{(6.626 × 10^{-34})(3 × 10^8)}{ 3300 × 10^{-10}}J\)

\(\implies E = \frac{(2 × 10^{-34})(3 × 10^8)}{ 10^{-10}} J\)

⇒ E = 6× 10 (-34 + 8 - 10) J = 6× 10 -19 Joule

So, the correct option is6× 10-19Joule

Concept:

• Bohr’s theory: He proposed that the electrons orbited in a specific orbit with a fixed radius of shells, with a fixed radius.
  • According to Bohr's model, the electron would absorb energy in the form of photons to get excited to a higher energy level.
  • It derives a radius for the nth excited state of hydrogen-like atoms.

The radius of Bohr's orbit

\(Z = \frac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}}} \times \frac{1}{Z}\)

Where n = principal quantum number of orbit, Z = atomic number, h = Plank’s constant, m = Mass of electron, e = elementary charge

Explanation:

• From the above discussion, it is clear that the r is directly proportional to n2.

From Bohr’s relation, r ∝ n2

So the correct option isn2/ Z.

CONCEPT:

• The photoelectric effect led to the conclusion that light is made up of a quantum of energy. Einstein through his experiments predicted that quanta of light have momentum which indicated the particle nature of light. This particle nature of light is called photons
• The properties of a photon are
  • Photon is an elementary particle
  • Irrespective of the frequency of radiation the momentum and energy of the photons remainsconstant for each photon
  • Photons are electrically neutral
  • The rest mass of the photon is zero and is electrically neutral

EXPLANATION:

• The pressure exerted by an electromagnetic wave is given by

\(\Rightarrow P = \frac{U}{C}\)

Where U = intensity of the wave, and C = Velocity of light

• The velocity of a photon is equal to the velocity of light, hence according to the above equation the photon must exert pressure on the surface
• Option 4 is the wrong statement.

Concept:

Bohr Model of Hydrogen Atom

• Bohr proposed a model for the hydrogen atomin which asingle electron revolves around a stationary nucleus of positive charge Ze(called hydrogen-like atom).
• A moving electron in its circular orbit behaves like a particle-wave.
• As a result, standing waves are produced and thetotal distance traveled by a wave is an integral number of wavelengths.

​This gives the relation:\(2πr_k = \frac{kh}{mv_k}\)= kλ ----(1)

Where rkis the radius of the kthorbit, andλ is the wavelength.

Also Thede Broglie wavelengthis given by:

λ =\(\frac{h}{mv_k}\) ----(2)

​Where vkis the velocity of the electron in kthorbit.

From (1) and (2) we get:

\(2πr_k = \frac{kh}{mv_k}\)

⇒\(mv_kr_k =\frac{kh}{2π}\)⇒\(m\omega_k =\frac{kh}{2π}\)

∴Angular momentum=\(m\omega_k =\frac{kh}{2π}\)

Explanation:

So, the angular momentum is given by

\(L =m\omega_k = mvr =\frac{kh}{2π}\)

k and h are constant

So, we can say that

L∝ h / 2π

So,\(\frac h {2\pi}\)is the correct option.

Concept:

Matter waves and De Broglie Wavelength:

• Matter waves: The wave associated with particles in motion is called matter waves.
• Louis de Broglie, a famousFrench physicist,proposed that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
• The wavelength of material waves isalso known as thede Broglie wavelength.It is given as

\(\lambda = \frac{h}{p}\)

h is Planck's constant, p is momentum

This equation is also known asde Broglie Equation.

Explanation:

• For a body in motion, the momentum is given as p = mv
• m is mass, v is the velocity
• So, the de Broglie equation in terms of m and v can be written as

\(\lambda = \frac{h}{mv}\)