The kinetic energy of electron and photon arc the same. The relation between their De - Broglie wavelengths λ e and λ p is

Question

The kinetic energy of electron and photon arc the same. The relation between their De - Broglie wavelengths λ e and λ p is (A)  λe = λp  (B)  λe < λp  (C)  λe > λp  (D)  λe =2 λp

Tardigrade Admin · Accepted Answer

De-broglie wavelength of electron, λe=(h/√2meke) De-broglie wavelength of photon, λp=(hc/kp) (λe/λp)=(Kp/c √2meKe) =√(K/2me c2)> 1 (∵ Ke=Kp=K) ∴ λe > λp

Q. The kinetic energy of electron and photon arc the same. The relation between their De - Broglie wavelengths $λ_{e}$ and $λ_{p}$ is

$λ_{e} = λ_{p}$

$λ_{e} < λ_{p}$

$λ_{e} > λ_{p}$

$λ_{e} = 2 λ_{p}$

De-broglie wavelength of electron,
$λ_{e} = \frac{h}{2 m _{e} k _{e}}$
De-broglie wavelength of photon, $λ_{p} = \frac{h c}{k _{p}}$
$\frac{λ _{e}}{λ _{p}} = \frac{K _{p}}{c 2 m _{e} K _{e}}$
$= \frac{K}{2 m _{e} c ^{2}} > 1$ $(∵ K_{e} = K_{p} = K)$
$∴ λ_{e} > λ_{p}$

Q. The kinetic energy of electron and photon arc the same. The relation between their De - Broglie wavelengths λe​ and λp​ is

λe​=λp​

λe​<λp​

λe​>λp​

λe​=2λp​