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de Broglie (1924) predicted that small p...

de Broglie (1924) predicted that small particles such as electrons should show wave -like properties along with paticle character. The wave length`(lamda)` associated with particle of mass m and moving with velocity v is given as `lamda=h/(mv)` where 'h' is plank's constant. The wave nature was confirmed by Davisson and Germer's experiment and modified equation for calculation of `lamda` can be given as:
`lamda=-h/(sqrt(2Em))` where E= kinetic energy of particle.
`lamda=h/(sqrt(2dVm))`, where d= change of particle accelerated potnetial of V volt.
If the kinetic energy of free electron is doubled,. Its de Broglie wavelengthh changes by the

A

`sqrt(2)`

B

`1/(sqrt(2))`

C

`2`

D

`1/2`

Text Solution

Verified by Experts

The correct Answer is:
B
`K.E.=1/2 mv^(2)=E`
`:.v=sqrt((2xxE)/m)`
`:.(v_(2))/(v_(1))=sqrt(2)`
Also `lamda_(1)=1/(mv_(1)),lamda_(2)=1/(mv_(2))`
`(lamda_(2))/(lamda_(1))=(v_(1))/(v_(2))implieslamda_(2)=((v_(1))/(v_(2)))xxlamda_(1)=1/(sqrt(2))xxlamda_(1)`
Hence b is the correct answer.
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de Broglie (1924) predicted that small particles such as electrons should show wave -like properties along with paticle character. The wave length (lamda) associated with particle of mass m and moving with velocity v is given as lamda=h/(mv) where 'h' is plank's constant. The wave nature was confirmed by Davisson and Germer's experiment and modified equation for calculation of lamda can be given as: lamda=-h/(sqrt(2Em)) where E= kinetic energy of particle. lamda=h/(sqrt(2dVm)) , where d= change of particle accelerated potnetial of V volt. Velocity of d e-Broglie's waves is given by

de Broglie (1924) predicted that small particles such as electrons should show wave -like properties along with paticle character. The wave length (lamda) associated with particle of mass m and moving with velocity v is given as lamda=h/(mv) where 'h' is plank's constant. The wave nature was confirmed by Davisson and Germer's experiment and modified equation for calculation of lamda can be given as: lamda=-h/(sqrt(2Em)) where E= kinetic energy of particle. lamda=h/(sqrt(2dVm)) , where d= change of particle accelerated potnetial of V volt. The ratio of de-Broglie chi wavelength of molecules of H_(2) and He at 27^(@)C adn 127^(@)C respectively is

Knowledge Check

  • A sine wave has an amplitude A and wavelength lamda . The ratio of particle velocity and the wave velocity is equal to (2piA=lamda)

    A
    `le1`
    B
    `=1`
    C
    `ge1`
    D
    data insufficient

  • The de Broglie waves are associated with moving particles. These particle may be

    A
    electrons
    B
    `He^(+), Li^(2+)` ions
    C
    Cricket ball
    D
    All of the above

  • The de-Broglie wavelength of a particle of charge q and mass m is lamda . If its kinetic energy be K, then

    A
    `lamda=(h)/(sqrt(mK))`
    B
    `lamda=(h)/(sqrt(mqK))`
    C
    `lamda=(sqrt(2mK))/(h)`
    D
    `lamda=(sqrt(mqK))/(h)`.

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