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

     

(none)

According to de-Broglie hypothesis, λ=hmv

mvr=nhπ  mv=nh2πr

λ=2πrhnh  λrn

λ1λ4=r1n4n1r4=5.3×1011×41×84.8×1011

 λ1λ4=14



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=6.63×10341030×2.21×1011

=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.