The de Broglie Wavelength of a molecule of thermal energy KT (K is Boltzmann constant. and T is absolute temperature) is given by

Question

The de Broglie Wavelength of a molecule of thermal energy KT (K is Boltzmann constant. and T is absolute temperature) is given by (A) (h/√2mKT) (B) (h/2mKT) (C) h√2mKT (D) (h/h√2mKT)

Tardigrade Admin · Accepted Answer

(1/2) mv 2= KT mv =√2 mKT So de-broglie wavelength λ=( h / mv ) arrow λ=( h /√2 mKT )

Q. The de Broglie Wavelength of a molecule of thermal energy KT (K is Boltzmann constant. and T is absolute temperature) is given by

$\frac{h}{2 m K T}$

$\frac{h}{2 m K T}$

$h 2 m K T$

$\frac{h}{h 2 m K T}$

$\frac{1}{2} m v^{2} = K T$
$m v = 2 m K T$
So de-broglie wavelength
$λ = \frac{h}{m v}$
$\to λ = \frac{h}{2 m K T}$

Q. The de Broglie Wavelength of a molecule of thermal energy KT (K is Boltzmann constant. and T is absolute temperature) is given by

2mKT​h​

2mKTh​

h2mKT​

h2mKT​h​

21​mv2=KT mv=2mKT​ So de-broglie wavelength λ=mvh​ →λ=2mKT​h​

$\frac{h}{2 m K T}$

$\frac{h}{2 m K T}$

$h 2 m K T$

$\frac{h}{h 2 m K T}$

$\frac{1}{2} m v^{2} = K T$
$m v = 2 m K T$
So de-broglie wavelength
$λ = \frac{h}{m v}$
$\to λ = \frac{h}{2 m K T}$