Q 1 :

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is

(Assume h=6.63×10-34 Js,me=9.0×10-31 kg and mp=1836 times me)          [2024]

  • 1:11836

     

  • 1:1836

     

  • 1:1836

     

  • 1:11836

     

(2)         

                λ is same for both, λp=λe

                P=hλ same for both

                 P=2mK

                Hence, K1mKEpKEe=memp=11836

 



Q 2 :

A proton and an electron have the same de Broglie wavelength. If Kp and Ke be the kinetic energies of proton and electron respectively, then choose the correct relation               [2024]

  • Kp=Ke2

     

  • Kp=Ke

     

  • Kp>Ke

     

  • Kp<Ke

     

(4)          

              De Broglie wavelength of proton and electron = λ

                λ=hppproton=pelectron

                 KE=p22mKE1m

                 KEeKEp=mpme

 



Q 3 :

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as            [2024]

  • λe>λα>λp

     

  • λα<λp<λe

     

  • λp<λe<λα

     

  • λp>λe>λα

     

(2)         

            λ=hP=h2mKE

              KE is same for all.λ1m

               me<mp<mα or λe>λp>λα



Q 4 :

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 25 % of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be             [2024]

  • 1/1

     

  • 1/8

     

  • 8/1

     

  • 1/4

     

(2)           

              For photon, Ep=hcλpλp=hcEp

                For electron, λe=hmeve=hve2Ke

                Given ve=0.25c

                           λe=h×0.25c2Ke=hc8Ke

                 Also hcEp=hc8KeKeEp=18

 



Q 5 :

The de Broglie wavelengths of a proton and an α particle are λ and 2λ respectively. The ratio of the velocities of proton and α particle will be           [2024]

  • 1 : 8

     

  • 1 : 2

     

  • 4 : 1

     

  • 8 : 1

     

(4)       

            λ=hp=hmvv=hmλ

              vpvα=mαmp×λαλp=4×2=8

 



Q 6 :

An electron in the ground state of the hydrogen atom has the orbital radius of 5.3×1011 m while that for the electron in third excited state is 8.48×1010 m. The ratio of the de-Broglie wavelengths of electron in the ground state to that in excited state is          [2025]

  • 4

     

  • 9

     

  • 3

     

  • 16

     

(1)

According to de-Broglie hypothesis, λ=hmv

mvr=nh2π  mv=nh2πr

λ=2πrhnh  λrn

λ1λ4=r1n4n1r4=5.3×1011×41×84.8×1011

 λ1λ4=14

Note: Most appropriate answer will be option (1).



Q 7 :

A sub-atomic particle of mass 1030 kg is moving with a velocity 2.21×106 m/s. Under the matter wave consideration, the particle will behave closely like __________.

(h=6.63×1034 Js)          [2025]

  • Infra-red radiation

     

  • X-ray

     

  • Gamma rays

     

  • Visible radiation

     

(2)

λ=hp=6.63×10341030×2.21×106

=3×1010 m=3A  X-ray

Hence, particle will behave as X-ray.



Q 8 :

An electron of mass 'm' with an initial velocity v=v0i^(v0>0) enters an electric field E=E0k^. If the initial de-Broglie wavelength is λ0, then its value after time t would be         [2025]

  • λ01+e2E02t2m2v02

     

  • λ01e2E02t2m2v02

     

  • λ0

     

  • λ01+e2E02t2m2v02

     

(1)

v=v0i^E0emtk^

|v|=v02+E02e2t2m2

λ0=hmv0

λ'=hmv01+E02e2t2v02m2

λ'=λ01+E02e2t2v02m2



Q 9 :

A proton of mass 'mp' has same energy as that of a photon of wavelength 'λ'. If the proton is moving at non-relativistic speed, then ratio of its de-Broglie wavelength to the wavelength of photon is.          [2025]

  • 1c2Emp

     

  • 1cEmp

     

  • 1cE2mp

     

  • 12cEmp

     

(3)

E is missing in the question but considering E as energy.

Energy of photon, Ephoton=hcλphoton

 Wavelength of photon, λphoton=hcE

Energy of proton, Eproton=12mpv2=p22mp

 Linear momentum of proton, p=2mpE

or de-Broglie wavelength of proton, λproton=hp=h2mpE

Ratio λprotonλphoton=h2mpE×Ehc=1cE2mp



Q 10 :

If λ and K are de-Broglie Wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be           [2025]

  •  

  •  

  •  

  •  

(2)

λ=hmv=h2mK

λ2=h22m(1k)

Y=cx2

Upward facing parabola passing through origin.



Q 11 :

An electron with mass 'm' with an initial velocity (t=0)v=v0i^(v0>0) enters a magnetic field B=B0j^. If the initial de-Broglie wavelength at t = 0 is λ0 then its value after time 't' would be:          [2025]

  • λ01e2B02t2m2

     

  • λ01+e2B02t2m2

     

  • λ01+e2B02t2m2

     

  • λ0

     

(4)

Speed does not change due to magnetic field, that's why λdebroglie remains unchanged.



Q 12 :

A photo-emissive substance is illuminated with a radiation of wavelength λi so that it releases electrons with de-Broglie wavelength λe. The longest wavelength of radiation that can emit photoelectron is λ0. Expression for de-Broglie wavelength is given by:

(m : mass of the electron, h : Planck's constant and c : speed of light)          [2025]

  • λe=h2mc(1λi1λ0)

     

  • λe=hλ02mc

     

  • λe=h2mc(1λi1λ0)

     

  • λe=hλi2mc

     

(1)

E=ϕ+KE  hcλe=ϕ+KE

λe=h2mK.E, E=hcλi and ϕ=hcλ0

h22mλe2=hcλihcλ0  λe=h2mc(1λi1λ0)



Q 13 :

An electron is released from rest near an infinite non-conducting sheet of uniform charge density 'σ'. The rate of change of de-Broglie wave length associated with the electron varies inversely as nth power of time. The numerical value of n is ________.          [2025]



(2)

Let the momentum of e at any time t is p and its de-broglie wavelength is λ.

Then, p=hλ

dpdt=hλ2dλdt

ma=F=hλ2dλdt   [m = mass of e]

Where –ve sign represents decrease in λ with time

a=p2mhdλdt=mv2hdλdt

dλdt=ahmv2          ... (i)

Here, a=qEm=emσ2ε0=σe2mε0

and v = u + at

u = 0, v = at

Substituting values of a and v in equation (i)

dλdt=2hε0σet2

 dλdt1t2  n=2



Q 14 :

An α-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength          [2023]

  • λα>λp>λe

     

  • λα<λp<λe

     

  • λα=λp=λe

     

  • λα>λp<λe

     

(2)

λD=hp=h2mK

 λ1m

 mα>mp>me

λe>λp>λα



Q 15 :

Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of λ0. If the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be                 [2023]

  • λ02

     

  • 3λ0

     

  • λ02

     

  • 9λ0

     

(3)

λ=h2mVq

λλ0=2040  λ=λ02



Q 16 :

The ratio of de-Broglie wavelength of an α-particle and a proton accelerated from rest by the same potential is 1m, the value of m is                   [2023]

  • 4

     

  • 16

     

  • 8

     

  • 2

     

(3)

λαλp=h2mαqαV=h2mpqpV

λαλp=18  m=8



Q 17 :

An electron accelerated through a potential difference V1 has a de-Broglie wavelength of λ. When the potential is changed to V2, its de-Broglie wavelength increases by 50%. The value of (V1V2) is equal to                [2023]

  • 3

     

  • 94

     

  • 32

     

  • 4

     

(2)

K.E.=P22m, P=hλ

eV1=(hλ)22m

eV2=(h1.5λ)22m

V1V2=(1.5)2=94



Q 18 :

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.            [2023]

Assertion A: The beam of electrons shows wave nature and exhibit interference and diffraction.

Reason R: Davisson Germer experimentally verified the wave nature of electrons.

In the light of the above statements, choose the most appropriate answer from the options given below:

  • A is correct but R is not correct

     

  • A is not correct but R is correct

     

  • Both A and R are correct but R is not the correct explanation of A

     

  • Both A and R are correct and R is the correct explanation of A

     

(4)

Beam of electrons show wave nature and exhibit interference and diffraction as shown by Davisson Germer experiment.



Q 19 :

A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of λ. An alpha particle having certain kinetic energy has the same de-Broglie wavelength λ. The ratio of kinetic energy of proton and that of alpha particle is                    [2023]

  • 2 : 1

     

  • 4 : 1

     

  • 1 : 2

     

  • 1 : 4

     

(2)

KE=p22m=h22mλ2

KEpKEα=mαmp=4:1



Q 20 :

The kinetic energy of an electron, α-particle and a proton are given as 4K, 2K and K respectively. The de-Broglie wavelength associated with electron (λe), α-particle (λα) and the proton (λp) are as follows                        [2023]

  • λα=λp>λe

     

  • λα>λp>λe

     

  • λα<λp<λe

     

  • λα=λp<λe

     

(3)

  Electron Alpha Proton
Mass: m1840 4m  m
Charge: e 2e e
Kinetic energy: 4K 2K K
λ=h2mK h2·(m1840)·4K h2.4m.2K h2mK

 

λα<λp<λe



Q 21 :

Proton (P) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, mP=1849me)              [2023]

  • 1 : 43

     

  • 1 : 1

     

  • 43 : 1

     

  • 1 : 1849

     

(2)

De Broglie wavelength is λ=hmv

λp=λe  mpvp=meve  pp=pe



Q 22 :

The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is λ1. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes                  [2023]

  • 12λ1

     

  • 2λ1

     

  • 12λ1

     

  • 2λ1

     

(1)

From K.T.G

vRMS=3kBTm

vRMST  and  hmvRMS=λ  λ1T

λ2λ1=T1T2=300600=12

  λ2=λ12



Q 23 :

The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy (Assume mp=me×1849)                        [2023]

  • 1 : 43

     

  • 1 : 30

     

  • 1 : 62

     

  • 2 : 43

     

(1)

λ1m  λpλe=memp=1:43



Q 24 :

A proton and an α-particle are accelerated from rest by 2V and 4V potential, respectively. The ratio of their de-Broglie wavelength is:               [2023]

  • 4 : 1

     

  • 2 : 1

     

  • 8 : 1

     

  • 16 : 1

     

(1)

λ=hmv=h2mK=h2mqΔV

λαλp=mpVpqpmαVαqα

  λαλp=1×2×14×4×2=14

  λp:λα=4:1



Q 25 :

The de Broglie wavelength of an electron having kinetic energy E is λ. If the kinetic energy of electron becomes E4, then its de-Broglie wavelength will be          [2023]

  • λ2

     

  • λ2

     

  • 2λ

     

  • 2λ

     

(3)

λ=h2mE

λ'=h2m(E4)=2h2mE=2λ



Q 26 :

The de Broglie wavelength of an oxygen molecule at 27°C is x×10-12 m. The value of x is (take Planck’s constant =6.63×10-34 J.s, Boltzmann constant =1.38×10-23 J/K, mass of oxygen molecule =5.31×10-26 kg).              [2026]

  • 20

     

  • 30

     

  • 26

     

  • 24

     

(3)

λ=h2mK=h2m(32kT)

λ=h3mkT

=6.63×10-343×5.31×10-26×1.38×10-23×300

=2.58×10-11=25.8×10-12

So, x=26



Q 27 :

A particle having electric charge 3×10-19 C and mass 6×10-27 kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is α×10-12 m. The value of α is_______.

(Take Planck's constant =6.6×10-34 J·s)          [2026]



(10)

λ=h2mqV

λ=6.6×10-342×18×10-46×1.21

λ=10-11 m=10×10-12 m

α=10



Q 28 :

The ratio of de Broglie wavelength of a deuteron with kinetic energy E to that of an alpha particle with kinetic energy 2E is n:1. The value of n is ____.    [2026]

(Assume mass of proton = mass of neutron.)



(2)

λ=hmv=h2m·KE

λdλα=mα·KEαmd·KEd=4m·2E2m·E=2:1