The wavelength of a photon and de-Broglie wavelength an electron have the same value. Given that v is the speed of electron and c is the velocity of light. Ee, Ep is the kinetic energy of electron and energy of photon respectively while pe, ph is the momentum of electron and photon respectively. Then which of the following relation is correct?

Question

The wavelength of a photon and de-Broglie wavelength an electron have the same value. Given that v is the speed of electron and c is the velocity of light. Ee, Ep is the kinetic energy of electron and energy of photon respectively while pe, ph is the momentum of electron and photon respectively. Then which of the following relation is correct? (A) (Ee/Ep)=(v/2 c) (B) (Ee/Ep)=(2 c/v) (C) (pe/ph)=(c/2 v) (D) (pe/ph)=(2 c/v)

Tardigrade Admin · Accepted Answer

Ee=(1/2)mv2=(1/2)(m v)v=(1/2)((h/λ ))v and Ep=(h c/λ ); ∴ (Ee/Ep)=(v/2 c) and according to De Broglie's equation ph=(h/λ ) ∴ (pe/ph)=1

$\frac{E _{e}}{E _{p}} = \frac{v}{2 c}$

$\frac{E _{e}}{E _{p}} = \frac{2 c}{v}$

$\frac{p _{e}}{p _{h}} = \frac{c}{2 v}$

$\frac{p _{e}}{p _{h}} = \frac{2 c}{v}$

$E_{e} = \frac{1}{2} m v^{2} = \frac{1}{2} (m v) v = \frac{1}{2} (\frac{h}{λ}) v$
and $E_{p} = \frac{h c}{λ};$
$∴ \frac{E _{e}}{E _{p}} = \frac{v}{2 c}$
and according to De Broglie's equation
$p_{h} = \frac{h}{λ}$
$∴ \frac{p _{e}}{p _{h}} = 1$

Ep​Ee​​=2cv​

Ep​Ee​​=v2c​

ph​pe​​=2vc​

ph​pe​​=v2c​

Ee​=21​mv2=21​(mv)v=21​(λh​)v and Ep​=λhc​; ∴Ep​Ee​​=2cv​ and according to De Broglie's equation ph​=λh​ ∴ph​pe​​=1

$\frac{E _{e}}{E _{p}} = \frac{v}{2 c}$

$\frac{E _{e}}{E _{p}} = \frac{2 c}{v}$

$\frac{p _{e}}{p _{h}} = \frac{c}{2 v}$

$\frac{p _{e}}{p _{h}} = \frac{2 c}{v}$

$E_{e} = \frac{1}{2} m v^{2} = \frac{1}{2} (m v) v = \frac{1}{2} (\frac{h}{λ}) v$
and $E_{p} = \frac{h c}{λ};$
$∴ \frac{E _{e}}{E _{p}} = \frac{v}{2 c}$
and according to De Broglie's equation
$p_{h} = \frac{h}{λ}$
$∴ \frac{p _{e}}{p _{h}} = 1$