Question

Solve this

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Answer 1

$\large\underline{\sf{Solution-}}$

We know,

Kinetic energy of a particle having mass m and velocity v is given by

$\rm :\longmapsto\:KE = \dfrac{1}{2} {mv}^{2} - - - - (1)$

Also,

Kinetic energy of a particle having absolute temperature T having Boltman's constant is

$\rm :\longmapsto\:KE = \dfrac{3}{2} KT - - - - (2)$

So, from equation (1) and (2), we get

$\rm :\longmapsto\:\dfrac{1}{2} {mv}^{2} = \dfrac{3}{2}KT$

$\rm :\longmapsto\: {mv}^{2} = 3KT$

can be rewritten as

$\rm :\longmapsto\: {m}^{2} {v}^{2} = 3KTm$

We know,

Momentum, p = mv

$\rm :\longmapsto\: {p}^{2} = 3KTm$

$\bf\implies \:p = \sqrt{3KTm} - - - (3)$

We know,

The relationship between wavelength and momentum is given by

$\rm :\longmapsto\:p = \dfrac{h}{\lambda}$

where, h is Planck's constant

$\rm :\longmapsto\: \sqrt{3KTm} = \dfrac{h}{\lambda}$

$\bf\implies \:\lambda \: = \: \dfrac{h}{ \sqrt{3KTm} }$

Hence, Proved.