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The de-Broglie wavelength of an electron...

The de-Broglie wavelength of an electron is same as the wavelength of a photon. The energy of a photon is ‘x’ times the K.E. of the electron, then ‘x’ is: (m-mass of electron, h-Planck’s constant, c - velocity of light)

A

`(hc)/(2lambdam)`

B

`(2lambdamc)/(h)`

C

`(2lambdac)/(hm)`

D

`(2lambdam)(ch)`

Text Solution

Verified by Experts

The correct Answer is:
B
Given `h/(mv) =lambda rArr v=h/(m lambda)`
Now, energy of photon, `E_(1)=(hc)/(lambda)`
And kinetic energy of electron `E_(2)=1/2mv^(2)`
`rArr E_(1)/E_(2)=(2hc)/(h) (mv)/(mv^(2)) rArr E_(1)/E_(2)=(2c)/(v)=(2cmlambda)/(h) `
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