Calculation of de-Broglie wavelength of neutrino, diatomic gas and poly-atomic gas

de-Broglie wavelength of neutrino:

Neutrino behaves as pont particle and has three degree of freedom associated with three independent velocity ( \(v_x, v_y, v_z\)).

So, K.E of thermal neutrino at temperature T

$$K.E= \frac f2 KT$$

$$= \frac32 KT$$

Where, K= Boltzmann constant \((1.38\times 10^{-23} j/k\))

$$or,\;\; \frac12 mv^2= \frac32 KT$$

$$or,\;\; \frac{p^2}{2m}=\frac{3}{2}KT$$

$$or,\;\; P=\sqrt{2mKT}\dotsm(1)$$

$$\lambda=\frac{h}{p}$$

$$=\frac{h}{\sqrt{3mkT}}\dotsm(2)$$

$$\Rightarrow \lambda \propto\frac{1}{\sqrt{T}}$$

$$\Rightarrow \lambda\propto\frac{1}{\sqrt{m}}$$

de-Broglie wavelength of monoatomic gas( He, Ar, Ne etc ):

Monoatomic gas behaves like point particle at given temperature and it has three degree of freedom.

$$\therefore \; K.E= \frac{p^2}{2m}= \frac32 KT$$

$$\therefore P= \sqrt{3mKT}$$

Hence, de-Broglie wave length is

$$\lambda=\frac hp=\frac{h}{\sqrt{3mKT}}\dotsm(1)$$

Example

Calculate the de-Broglie wavelength of He- atom at \(27^0 C\)

Solution:

T= ( 27 + 272 ) K

We know,

$$6.023\times 10^{23} atom= 4gm$$

$$1 atom = \frac{4}{6.023\times 10^{23}gm}$$

Now,

$$\lambda= \frac{6.626\times 10^{-34}}{\sqrt{3\times \frac{4}{6.023\times 10^{23}} \times 1.38\times 10^{-23}\times 300}}$$

$$= 2.31\times 10^{-12}m$$

$$=0.0231 A^0$$

de-Broglie wavelength of diatomic gas:

The diatomic molecule at given temperature T ( low - temperature ) behaves like a rigid dumb-bell. It has five degree of reedom associated with velocities (\( v_x, v_y, v_x\;and \; \omega_y\)) i.e 3- translational and 2- rotational degree of freedom.

$$K.E= \frac f2 KT$$

$$\frac{p^2}{2m}=\frac52 KT$$

$$p=\sqrt{5mKT}\dotsm(1)$$

$$\lambda=\frac{h}{p}$$

$$=\frac{h}{\sqrt{5mKT}}\dotsm(2)$$

This expression is for the toom temperature.

The diatomic molecule at given temperature T ( high- temperature ) behaves like spring dumbbell. It has 7-degree of freedom associated with velocities ( \(v_x,v_y ,v_z,\omega_x\) and \(\omega_y\)) and extension compression of bound (spring)

$$K.E= \frac f2 KT$$

$$\frac{p^2}{2m}=\frac72 KT$$

$$\therefore p= \sqrt{7mKT}\dotsm(3)$$

$$\therefore\;\; \lambda=\frac{h}{p}$$

$$=\frac{h}{\sqrt{7mKT}}$$

de-Broglie wavelength of polyatomic gas:

The degree of freedom for polyatomic gas molecule is taken as 6. So de-Broglie wavelength is given by

$$\lambda=\frac{h}{\sqrt{6mKT}}$$

In case of most probable distribution of velocity in Maxwell Boltzmann statistic. The de-Broglie wavelength is calclated by using most probable velocity.

$$\lambda=\frac{h}{mv_p}$$

$$=\frac{h}{m\sqrt{\frac{2KT}{m}}}$$

$$=\frac{h}{\sqrt{2mKT}}$$

Reference:

1. Mathews, P.M and K Venkatesan. A Text Book of Quantum Mechanics. New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
2. Merzbacher, E. Quantum Mechanics . New York: John Wiley, 1969.
3. Prakash, S and S Salauja. Quantum Mechanics. Kedar Nath Ram Nath Publishing Co, 2002.
4. Singh, S.P, M.K and K Singh. Quantum Mechanics. Chand & Company Ltd., 2002.