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))/(E_(e^(C)))=10^(-1)`
D. `(P_(e))/(E_(e^(C)))=10^(-4)`

Answer

B

Explanation

(b) : For electron,
`lamda_(e)=(h)/(m_(e)v_(e))=(h)/(m_(e)(c//100))=(100h)/(m_(e)c)` . . .(i)
`E_(e)=(1)/(2)m_(e)v_(e)^(2) or m_(e)v_(e)=sqrt(2E_(e)m_(e))`
`lamda_(e)=(h)/(m_(e)v_(e))=(h)/(sqrt2m_(e)E_(e)) or E_(e)=(h^(2))/(2lamda_(e)^(2)m_(e))` . . .(ii)
For photon, `E_(p)=(hc)/(lamda_(p))=(hc)/(2lamda_(e))" "( :.lamda_(p)=2lamda_(e)"(Given)"`
`:.(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`
`:.(E_(e))/(E_(p))=(1)/(100)=10^(-2)` (Using(i))
For electron, `p_(e)=m_(e)v_(e)=m_(e)xx(c)/(100)`
`:.(p_(e))/(m_(e)c)=(1)/(100)=10^(-2)` Thus, option (b) is correct.