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

C

Explanation

`{:("Suppose, Mass of electron"=m_(e),"Mass of photon"=m_(p)),("Velocity of electron"=v_(e)and,"Velecity of photon"=v_(p)):}`
Thus, for electronk, de-Broglie wavelength
`" "lamda_(e)=h/(m_(e)v_(e))`
`" "=h/(m_(e)(c//100))=(100h)/(m_(e)c)` (given)
kinetic energy, `" "E_(e)=1/2m_(e)v_(e)^(2)`
`implies" "m_(e)v_(e)=sqrt(2E_(e)m_(e))`
So, `" "lamda_(e)=h/(m_(e)v_(e))=h/(sqrt(2m_(e)E_(e)))`
`implies" "E_(e)=(h_(2))/(2lamda_(e)^(2)m_(e))`
For photon of wavelength `lamda_(p),` energy
`" "E_(p)=(hc)/(lamda_(p))=(hc)/(2lamda_(e))" "[becauselamda_(p)=2lamda_(e)]`
`therefore" "(E_(p))/(E_(e))=(hc)/(2lamda_(e))xx(2lamda_(e)^(2)m_(e))/(h^(2))`
`" "=(lamda_(e)m_(e)c)/(h)=(100h)/(m_(e)c)xx(m_(e)c)/(h)=100`
So, `" "(E_(e))/(E_(p))=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)`