The de Broglie wavelength of a molecule of thermal energy KT (K is Boltzmann constant and T is absolute temperature) is given by :
$\frac{h}{\sqrt{2 m K T}}$
$\frac{h}{2 m K T}$
$h \sqrt{2 m K T}$
$\frac{1}{h \sqrt{2 m K T}}$

Question

The de Broglie wavelength of a molecule of thermal energy KT -K is Boltzmann constant and T is absolute temperature- is given by -dfrac-h-sqrt-2mKT-dfrac-h-2mKT-hsqrt-2mKT-dfrac-1-hsqrt-2mKT-

Toppr Admin · Accepted Answer

12mv2-KT

The de Broglie wavelength of a molecule of thermal energy KT (K is Boltzmann constant and T is absolute temperature) is given by :h√2mKTh2mKTh√2mKT1h√2mKT

12mv2=KT

The de Broglie wavelength of a molecule of thermal energy KT (K is Boltzmann constant and T is absolute temperature) is given by :
$\frac{h}{\sqrt{2 m K T}}$
$\frac{h}{2 m K T}$
$h \sqrt{2 m K T}$
$\frac{1}{h \sqrt{2 m K T}}$

$\frac{1}{2} m v^{2} = K T$