Home
Class 12
PHYSICS
The deBroglie wavelength of a particle o...

The deBroglie wavelength of a particle of kinetic energy K is `lamda`. What would be the wavelength of the particle, if its kinetic energy were `(K)/(4)`?

Text Solution

Verified by Experts

If `lamda` is the de Broglie wavelength of the particle of kinetic energy K, then
`lamda=(h)/(sqrt(2mK))`
Suppose that the de Broglie wavlength of the particle becomes `lamda^``, when its kinetic energy is `(K)/(4)`, then
`lamda^`=(h)/(sqrt((2mK)/(4)))=2((h)/(sqrt(2mK)))=2lamda`
Promotional Banner

Topper's Solved these Questions

  • PHOTOELECTRIC EFFECT

    CENGAGE PHYSICS|Exercise Exercise 3.2|8 Videos
  • PHOTOELECTRIC EFFECT

    CENGAGE PHYSICS|Exercise Subjective|16 Videos
  • PHOTOELECTRIC EFFECT

    CENGAGE PHYSICS|Exercise Examples|8 Videos
  • NUCLEI

    CENGAGE PHYSICS|Exercise QUESTION BANK|59 Videos
  • RAY OPTICS

    CENGAGE PHYSICS|Exercise DPP 1.6|12 Videos

Similar Questions

Explore conceptually related problems

Debroglie wavelength of a particle at rest position is

Debroglie wavelength of uncharged particles depends on

The de Broglie wavelength (lambda) of a particle is related to its kinetic energy E as

The de-Broglie wavelength of a particle having kinetic energy E is lamda . How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value ?

Assertiom: The de-broglie wavelength of a neutrons when its kinetic energy is k is lambda . Its wavelength is 2lambda when its kinetic energy is 4k . Reason : The de-broglie wavelength lambda is proportional to square root of the kinetic energy.