Q: A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to

Velocity of a body falling from a height H is given byv=2gHWe know that de-broglie wavelengthλ=mVh=m2gHh⇒=m2gHhHere, m2gh is a constant ϕ say 'K'.So,λ=KH1⇒λ∝H1⇒λ∝H−1/2

Q: Consider a beam of electrons (each electron with energy E0 ) incident on a metal surface kept in an evacuated chamber. Then,

When a beam of electrons of energy E0 is incident on a metal surface kept in an evacuated chamber electrons can be emitted with maximum energy E0 (due to elastic collision) and with any energy less than E0. when part of incident energy of electron is used in liberating the electrons from the surface of metal.

Q: An electron is moving with an initial velocity v=v0i^ and is in a magnetic field B=B0j^. Then, it's de-Broglie wavelength

Given,v=v0i^⇒B=B0j^Force on moving electron due to magnetic field is, F=−e(v×B)=−e[v0i^×B0j^]⇒=−ev0B0kAs this force is perpendicular to v and B, so the magnitude of v will not change, i.e., momentum (=mv) will remain constant in magnitude. Hence,de-Broglie wavelength λ=mvh remains constant.

Q: An electron (mass m ) with an initial velocity v=v0i(v0>0) is in an electric field E=−E0i^(E0= constant >0). It's de-Broglie wavelength at time t is given by

Initial de-Broglie wavelength of electron,λ0=mv0hForce on electron in electric field,F=−eE=−e[−E0i^]=eE0i^Acceleration of electron a=mF=meE0i^Velocity of electron after time t,v=v0i^+(meE0i^)t=(v0+meE0)i^=v0(1+mv0eE0t)i^de - Broglie wavelength associated with electron at time t isλ=mvh=m[v0(1+mv0eE0)]h=[1+mv0eE0t]λ0[∵λ0=mv0h]

Q: An electron (mass m ) with an initial velocity v=v0i^ is in an electric field E=E0j^. If λ0=h/mv0, it's de-Broglie wavelength at time t is given by

Initial de - Broglie wavelength of electron,λ0=mv0hForce on electron in electric field,F=−eE=−eE0j^Acceleration of electron, a=mF=mϵE0j^It is acting along negative y-axis.The initial velocity of electron along x-axis, vx0=v0i^. Initial velocity of electron along y-axis, vy0=0.Velocity of electron after time t along x-axis, vx=v0i^( ∵ there is no electron along x-axis.)Velocity of electron after timet along y-axis,vy=0+(−meE0j^)t=−meE0tj^Magnitude of velocity of electron after time t isv=vx2+vy2=v02+(m−eE0t)2⇒=v01+m2v02e2E02t2 de-Broglie wavelength, λ′=mvh=mv01+e2E02t2/(m2v02)h=1+e2E02t2/m2v02λ0

Q: Relativistic corrections become necessary when the expression for the kinetic energy 21mv2, becomes comparable with mc2, where m is the mass of the particle. At what de-Broglie wavelength, will relativistic corrections become important for an electron?

de-Broglie wavelengthλ=mvh⇒v=mλhHere, h=6.6×10−34 Jsand for electron, m=9×10−31 kgNow consider each option one by one(a) λ1⇒v1⇒=10 nm=10×10−9 m=10−8 m=(9×10−31)×10−86.6×10−34=32.2×105=105 m/s(b) λ2=10−1 nm=10−1×10−9 m=10−10 m⇒v2=(9×10−31)×10−1066×10−34=107 m/s(c) λ3=10−4 nm=10−4×10−9 m=10−13 m⇒v3=(9×10−31)×10−1366×10−34=1010 ms(d) λ4=10−6 nm=10−6×10−9 m=10−15 m⇒v4=9×10−31×10−1566×10−34=1012 m/sThus, options (c) and (d) are correct as v3 and v4 is greater than 3×108 m/s.

Q: Two particles A1 and A2 of masses m1,m2(m1>m2) have the same de-Broglie wavelength. Then,

de-Broglie wavelength λ=mvh where, mv=p (imoment)⇒λ=ρh⇒p=λhHere, h is a constant.So,p∝λ1⇒p2p1=λ1λ2But (λ1=λ2)=λThen, p2p1=λλ=1⇒p1=p2Thus, their momenta is same.Also, E=21mv2=21mmv2×m=21mm2v2=21mp2Here, ρ is constantE∝m1∴rE1=m1m2<1⇒E1<E2

Question 1

A particle is dropped from a height H. The de-Broglie wavelength of the particle as a function of height is proportional to

Accepted Answer

Velocity of a body falling from a height H is given byv=2gH​We know that de-broglie wavelengthλ=mVh​=m2gH​h​⇒=m2g​H​h​Here, m2g​h​ is a constant ϕ say 'K'.So,λ=KH​1​⇒λ∝H​1​⇒λ∝H−1/2

Question 2

The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly

Accepted Answer

Given in the question,Energy of a photon, E=1 MeV=106 eVNow, hc=1240 eVnmNow, E=λhc​⇒l=Ehc​=106 eV1240 eVnm​=1.2×10−3 nm

Question 3

Consider a beam of electrons (each electron with energy E0​ ) incident on a metal surface kept in an evacuated chamber. Then,

Accepted Answer

When a beam of electrons of energy E0​ is incident on a metal surface kept in an evacuated chamber electrons can be emitted with maximum energy E0​ (due to elastic collision) and with any energy less than E0​. when part of incident energy of electron is used in liberating the electrons from the surface of metal.

Question 4

Consider figure given below. Suppose the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that

Accepted Answer

In Davisson-Germer experiment, the de-Broglie wavelength associated with electron isλ=V​12.27​A˚…(i)where V is the applied voltage.If there is a maxima of the diffracted electrons at an angle θ, then2dsinθ=λ…(ii)From Eq. (i), we note that if V is inversely proportional to the wavelength λ.i.e. V will increase with the decrease in the λ.From Eq. (ii), we note that wavelength λ is directly proportional to sinθ and hence θ.So, with the decrease in λ,θ will also decrease.Thus, when the voltage applied to A is increased. The diffracted beam will have the maximum at a value of θ that will be less than the earlier value.

Question 5

A proton, a neutron, an electron and an α-particle have same energy. Then, their de-Broglie wavelengths compare as

Accepted Answer

We know that the relation between λ and K is given byλ=2mkhHere, for the given value of energy K,2kh is a constant.Thus, λ∝m1∴⇒λp:λn:λp:λα=mp1:mn1:m01:mα1Since, mp=mn, hence λp=λnAs, mα>mp, therefore λα<λpAs, meλnHence, λα<λp=λn<λe

Question 6

An electron is moving with an initial velocity v=v0​i^ and is in a magnetic field B=B0​j^​. Then, it's de-Broglie wavelength

Accepted Answer

Given,v=v0​i^⇒B=B0​j^​Force on moving electron due to magnetic field is, F=−e(v×B)=−e[v0​i^×B0​j^​]⇒=−ev0​B0​kAs this force is perpendicular to v and B, so the magnitude of v will not change, i.e., momentum (=mv) will remain constant in magnitude. Hence,de-Broglie wavelength λ=mvh​ remains constant.

Question 7

An electron (mass m ) with an initial velocity v=v0​i(v0​>0) is in an electric field E=−E0​i^(E0​= constant >0). It's de-Broglie wavelength at time t is given by

Accepted Answer

Initial de-Broglie wavelength of electron,λ0​=mv0​h​Force on electron in electric field,F=−eE=−e[−E0​i^]=eE0​i^Acceleration of electron a=mF​=meE0​i^​Velocity of electron after time t,v​=v0​i^+(meE0​i^​)t=(v0​+meE0​​)i^=v0​(1+mv0​eE0​​t)i^​de - Broglie wavelength associated with electron at time t isλ​=mvh​=m[v0​(1+mv0​eE0​​)]h​=[1+mv0​eE0​​t]λ0​​[∵λ0​=mv0​h​]​

Question 8

An electron (mass m ) with an initial velocity v=v0​i^ is in an electric field E=E0​j^​. If λ0​=h/mv0​, it's de-Broglie wavelength at time t is given by

Accepted Answer

Initial de - Broglie wavelength of electron,λ0​=mv0​h​Force on electron in electric field,F=−eE=−eE0​j^​Acceleration of electron, a=mF​=mϵE0​j^​​It is acting along negative y-axis.The initial velocity of electron along x-axis, vx0​​=v0​i^. Initial velocity of electron along y-axis, vy0​​=0.Velocity of electron after time t along x-axis, vx​=v0​i^( ∵ there is no electron along x-axis.)Velocity of electron after timet along y-axis,vy​=0+(−meE0​​j^​)t=−meE0​​tj^​Magnitude of velocity of electron after time t isv=vx2​+vy2​​=v02​+(m−eE0​​t)2​⇒=v0​1+m2v02​e2E02​t2​​ de-Broglie wavelength, λ′=mvh​=mv0​1+e2E02​t2/(m2v02​)​h​=1+e2E02​t2/m2v02​​λ0​​​

Question 9

Relativistic corrections become necessary when the expression for the kinetic energy 21​mv2, becomes comparable with mc2, where m is the mass of the particle. At what de-Broglie wavelength, will relativistic corrections become important for an electron?

Accepted Answer

de-Broglie wavelengthλ=mvh​⇒v=mλh​Here, h=6.6×10−34 Jsand for electron, m=9×10−31 kgNow consider each option one by one(a) λ1​⇒v1​⇒​=10 nm=10×10−9 m=10−8 m=(9×10−31)×10−86.6×10−34​=32.2​×105=105 m/s​(b) λ2​=10−1 nm=10−1×10−9 m=10−10 m⇒v2​=(9×10−31)×10−1066×10−34​=107 m/s​(c) λ3​=10−4 nm=10−4×10−9 m=10−13 m⇒v3​=(9×10−31)×10−1366×10−34​=1010 ms​(d) λ4​=10−6 nm=10−6×10−9 m=10−15 m⇒v4​=9×10−31×10−1566×10−34​=1012 m/s​Thus, options (c) and (d) are correct as v3​ and v4​ is greater than 3×108 m/s.

Question 10

Two particles A1​ and A2​ of masses m1​,m2​(m1​>m2​) have the same de-Broglie wavelength. Then,

Accepted Answer

de-Broglie wavelength λ=mvh​ where, mv=p (imoment)⇒λ=ρh​⇒p=λh​Here, h is a constant.So,p∝λ1​⇒p2​p1​​=λ1​λ2​​But (λ1​=λ2​)=λThen, p2​p1​​=λλ​=1⇒p1​=p2​Thus, their momenta is same.Also, E​=21​mv2=21​mmv2×m​=21​mm2v2​=21​mp2​​Here, ρ is constantE∝m1​∴rE1​​=m1​m2​​<1⇒E1​<E2​

Question 11

The de-Broglie wavelength of a photon is twice, the de-Broglie wavelength of an electron. The speed of the electron is ve​=100c​. Then,

Accepted Answer

Suppose, Mass of electron =me​. Mass of photon =mρ​.Velocity of electron =ve​ and  Velocity of photon =vρ​Thus, for electron, de-Broglie wavelengthλe​​=me​ve​h​=me​(c/100)h​=me​c100h​ (given) ​Kinetic energy, Ee​=21​me​ve2​⇒ so, ⇒​me​ve​=2Ee​me​​λe​=me​ve​h​=2me​Ee​​h​Ee​=2λe2​me​h2​​For photon of wavelength λp​, energyEp​=λp​hc​=2λp​hc​[∵λp​=2λe​]Ee​Eρ​​Ep​Ee​​​=2λe​hc​×h22λe2​me​​=hλe​me​c​=me​c100h​×hme​c​=100=1001​=10−2​For electron, pe​=me​ve​=me​×c/100So, me​cpe​​=1001​=10−2

Question 12

Photons absorbed in matter are converted to heat. A source emitting n photon/sec of frequency v is used to convert 1 kg of ice at 0∘C to water at 0∘C. Then, the time T taken for the conversion

Accepted Answer

Energy spent to convert ice into water= mass × latent heat =m L=(1000 g)×(80cal/g)=80000cal​Energy of photons used =nT×E=nT×hvSo, nThv=mL⇒T=nhvmL​∴T∝n1​, when v is constant.T∝v1​, when n is fixed.⇒T∝nv1​Thus, T is constant, if nv is constant.

Question 13

A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de-Broglie wavelength of the particle varies cyclically between two values λ1​,λ2​ with λ1​>λ2​. Which of the following statement are true?

Accepted Answer

The de-Broglie wavelength of the particle can be varying cyclically between two values λ1​ and λ2​, if particle is moving in an elliptical orbit with origin as its one focus.Consider the figure given belowLet v1​,v2​ be the speed of particle at A and B respectively and origin is at focus O. If λ1​,λ2​ are the de-Broglie wavelengths associated with particle while moving at A and B respectively. Then,λ1​=mv1​h​λ2​=mv2​h​λ2​λ1​​=v1​v2​​λ1​>λ2​v2​>v1​​By law of conservation of angular momentum, the particle moves faster when it is closer to focus.From figure, we note that origin O is closed to P than A.

Question 14

A proton and an α-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λp​ and λα​ related to each other?

Accepted Answer

λλλα​λp​​λp​​=2mqv​h​∝mq​1​=mp​qp​​mα​qα​​​=mp​×e​4mp​×2e​​=8​=8​λα​​i.e. wavelength of proton is 8​ times wavelength of α-particle.

Question 15

(i) In the explanation of photoelectric effect, we assume one phot frequency v collides with an electron and transfers its energy. leads to the equation for the maximum energy Emax​ of the en electron asEmax​=hv−ϕ0​where ϕ0​ is the work function of the metal. If an electron absorbs 2 photons (each of frequency v ), what will be the maximum ener the emitted electron?(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?

Accepted Answer

(i) Here it is given that, an electron absorbs 2 photons each of frequency v then v′=2v where, v′ is the frequency of emitted electron.Given, Emax​=hv−ϕ0​Now, maximum energy for emitted electrons isEmax′​=h(2v)−ϕ0​=2hv−ϕ0​(ii) The probability of absorbing 2 photons by the same electron is very law. Hence, such emission will be negligible.

Question 16

There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength.

Accepted Answer

According to first statement, when the materials which absorb photons of shorter wavelength has the energy of the incident photon on the material is high and the energy of emitted photon is law when it has a longer wavelength.But in second statement, the energy of the incident photon is law for the substances which has to absorb photons of larger wavelength and energy of emitted photon is high to emit light of shorter wavelength. This means in this statement material has to supply the energy for the emission of photons.But this is not possible for a stable substances.

Question 17

Do all the electrons that absorb a photon come out as photoelectrons?

Accepted Answer

In photoelectric effect, we can observe that most electrons get scattered into the metal by absorbing a photon.Thus, all the electrons that absorb a photon doesn't come out as photoelectron. Only a few come out of metal whose energy becomes greater than the work function of metal.

Question 18

There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength?

Accepted Answer

Suppose wavelength of X-rays is λ1​ and the wavelength of visible light is λ2​.Pλ1​λ2​​=100 W=1 nm=500 nm​Also, n1​ and n2​ represents number of photons of X-rays and visible light emitted from the two sources per sec.tE​=P=n1​λ1​hc​=n2​λ2​hc​λ1​n1​​=λ2​n2​​n2​n1​​=λ2​λ1​​=5001​​

Question 19

Consider figure for photoemission. How would you reconcile with momentum-conservation? Note light (photons) have momentum in a different direction than the emitted electrons.

Accepted Answer

During photoelectric emission, the momentum of incident photon is transferred to the metal. At microscopic level, atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.The excited electron is emitted. Therefore, the conservation of momentum is to be considered as the momentum of incident photon transferred to the nucleus and electrons.

Question 20

Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

Accepted Answer

Given,For the first condition,Wavelength of light λ=600 nmand for the second condition,Wavelength of light λ′=400 nmAlso, maximum kinetic energy for the second condition is equal to the twice of the kinetic energy in first condition.Kmax​Kmax​2Kmax​2(6001230​−ϕ)ϕ​=2Kmax​=λhc​−ϕ=λ′hc​−ϕ0​=(4001230​−ϕ)[∵hc=1240 Vnm]=12001230​=1.02eV​

Chapter 11 : Dual Nature of Radiation and Matter