In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage V then de-Broglie wavelength of that electron will be m.

Question

In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage V then de-Broglie wavelength of that electron will be m. (A) (2 V e m/√h) (B) (√h/2 V e m) (C) (√2 V e m/h) (D) (h/√2 V e m)

Tardigrade Admin · Accepted Answer

Kinetic energy (1/2)mv2=eV v=√(2 e V/m) Wavelength λ =(h/m v) λ =(h/m √(2 e V)m)=(h/√2 e V m)

Q. In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage V then de-Broglie wavelength of that electron will be ______m.

$\frac{2 V e m}{h}$

$\frac{h}{2 V e m}$

$\frac{2 V e m}{h}$

$\frac{h}{2 V e m}$

Kinetic energy $\frac{1}{2} m v^{2} = e V$
$v = \frac{2 e V}{m}$
Wavelength $λ = \frac{h}{m v}$
$λ = \frac{h}{m \frac{2 e V}{m}} = \frac{h}{2 e Vm}$

Q. In experiment of Davisson-Germer, emitted electron from filament is accelerated through voltage V then de-Broglie wavelength of that electron will be ______m.

h​2Vem​

2Vemh​​

h2Vem​​

2Vem​h​

Kinetic energy 21​mv2=eV v=m2eV​​ Wavelength λ=mvh​ λ=mm2eV​​h​=2eVm​h​

$\frac{2 V e m}{h}$

$\frac{h}{2 V e m}$

$\frac{2 V e m}{h}$

$\frac{h}{2 V e m}$

Kinetic energy $\frac{1}{2} m v^{2} = e V$
$v = \frac{2 e V}{m}$
Wavelength $λ = \frac{h}{m v}$
$λ = \frac{h}{m \frac{2 e V}{m}} = \frac{h}{2 e Vm}$