Question: Q. 1. An α-particle and a proton are accelerated through the same potential. Find the ratio of their de-Broglie wavelengths.

U] [Delhi 2017]

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Solution:

Ans. de-Broglie wavelength is given by

λ=h2mK

Mass of a proton =u

Charge of proton =e

Hence, Kinetic energy of a proton when accelerated through a potential =eV

Putting these values in de-Broglie wavelength formula to calculate wave length of proton

λp=h2ueV

Now,

  • Mass of an α-particle =4u
  • Charge of an α-particle =2e

Hence, Kinetic energy of an α-particle when accelerated through a potential =2eV

Putting these values in de-Broglie wavelength formula to calculate wave length of α-particle

λα=h8u×2eV=h16ueV

Hence ratio of λpλα=18

λpλα=122

[CBSE Marking Scheme 2017]



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