The de-Broglie wavelength of a photon and electron are same. Ee, Ep are the kinetic energy of the electron and photon, respectively. pe, ph are the momentum of electron and photon, respectively. Then, which of the following relations are correct? (.v= textspeed of electron,c= textvelocity of light.)

Question

The de-Broglie wavelength of a photon and electron are same. Ee, Ep are the kinetic energy of the electron and photon, respectively. pe, ph are the momentum of electron and photon, respectively. Then, which of the following relations are correct? (.v= textspeed of electron,c= textvelocity of light.) (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) 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 $E_{p} = \frac{h c}{λ}$
$∴ \frac{E _{e}}{E _{p}} = \frac{v}{2 c}$
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​ 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 $E_{p} = \frac{h c}{λ}$
$∴ \frac{E _{e}}{E _{p}} = \frac{v}{2 c}$
According to de-Broglie's equation,
$p_{h} = \frac{h}{λ}$
$∴ \frac{p _{e}}{p _{h}} = 1$