Home
Class 12
PHYSICS
A photon and an electron moving with a ...

A photon and an electron moving with a velocity v have the same de broglie wavelength . Then the ratio of the kinetic energy of the electron to the kinetic energy of the photon is [C is the speed of light ]

A

`v/(3c)`

B

`(3c)/v`

C

`(2c)/v`

D

`v/(2c)`

Text Solution

Verified by Experts

The correct Answer is:
D
`lambda_(e ) = lambda_(ph)`
`h/(P^(e))=h/(P_(ph))`
`sqrt(2mk_(e))=(E_(ph))/c`
`2mk_(e) = ((E_(ph))^(2))/(c^(2))`
`(k_(e))/(E_(ph))=(E_(ph))/(c^(2))(1/(2m))`
=`(p_(ph))/c(1/(2m))`
`=(p_(e))/c (1/(2m))`
`= (mv)/c 1/(2m)`
`=v/(2c)` .
Promotional Banner

Similar Questions

Explore conceptually related problems

An electron and proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

An electron and a photon have the same de Broglie wavelength. Which one of these has higher kinetic energy?

Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV.

The de-Broglie wavelength of an electron moving with a velocity 1.5 xx 10^(8)ms^(-1) is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the photon is:

The de broglie wavelength of electron moving with kinetic energy of 144 eV is nearly

An electron and a photon have got the same de- Broglie wavelength. Prove that total energy of electron is greater than energy of photon.

The de - Broglie wavelength of a particle moving with a velocity 2.25 xx 10^(8) m//s is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is (velocity of light is 3 xx 10^(8) m//s