The de-broglie wavelength of a photon is twice the de-broglie wavelength of an electron. The speed of the electron is `v_(e)=c/100`. Then

Options

A. `E_(e)/E_(p)=10^(-4)`
B. `E_(e)/E_(p)=10^(-2)`
C. `P_(e)/(m_(e)C)=10^(-2)`
D. `P_(e)/(m_(e)C)=10^(-4)`

Answer

B, C

Explanation

Suppose, Mass of electron =`m_(e)`,
Mass of photon =`m_(p)`
Velocity of electron =`v_(e)` and Velocity of photon =`v_(p)`
Thus,for electron, de-Broglie wavelength `lambda_(e)=h/(m_(e)v_(e))=h/(m_(e)(C//100))=(100h)/(m_(e)C)`(given)
Kinetic energy, `E_(0)=1/2m_(e)v_(e)^(2)`
`rArr m_(e)v_(e)=sqrt(2m_(e)m_(e))`
so,`lambda_(e)=h/(m_(e)v_(e))=h/sqrt(2m_(e)E_(e))`
`rArr E_(e)=h^(2)/(2lambda_(e)^(2)m_(e))`
For photon of wavelength `lambda_(p)`, energy
`E_(p)=(hc)/lambda_(p)=(hc)/(2lambda_(e))`
`therefore E_(p)/E_(e)=(hc)/(2lambda_(e))xx(2lambda_(e)^(2)m_(e))/h^(2)`
`=(lambda_(e)m_(e)C)/h=(100h)/(m_(e)c)xx(m_(e)C)/h=100`
So, `E_(p)/E_(e)=1/100=10^(-2)`
For electron, `p_(e)=m_(e)v_(e)=m_(e)xxc//100`
So,`p_(e)/(m_(e)c)=1/100=10^(-2)`