de-Broglie Wavelength of relativistic particle

de-Broglie Wavelength of relativistic particle :

A particle which is moving with high speed is associated with K.E and rest mass energy. The total energy of moving particle is given by

E= K.E + rest mass energy

E= \(E_k+m_0 c^2\dotsm(1)\)

the relation between total energy of particle and relativistic momentum is given by,

$$E=\sqrt{p^2c^2+m_0^2 c^4}\dotsm(2)$$

Where,

\(m_0\)= Rest mass

C= Speed of light

P= linear momentum

From equation (1) and (2)

$$\sqrt{p^2 c^2+ m_0^2 c^4}= E_k+ m_0 c^2$$

Squaring both side

$$p^2c^2 + m_0^2 c^4= E_k^2+ 2E_k m_0 c^2 + m_0^2 c^4$$

$$P=\sqrt{\frac{E_k^2 + 2E_k m_0 c^2}{c^2}}$$

$$or,\;\; \sqrt{2m_0 E_k}\biggl[1+ \frac{E_k}{2m_0 c^2}\biggr]^{\frac12}\dotsm(3)$$

Here, equation (3) gives relativistic linear momentum of particle.

The de-Broglie wavelength of particle is given by,

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

$$= \frac{h}{\sqrt{2m_0E_k}}\biggl[1+ \frac{E_k}{2m_0 c^2}\biggr]^{-\frac12}\dotsm(4)$$

Comparing with non-relativistic particles

$$\lambda=\frac{h}{\sqrt{2m_0 E_k}}\dotsm(5)$$

In equation (4) \(\biggl(+\frac{E_k}{2m_0 c^2}\biggr)^{-\frac12}\) is known as relativistic correction factor for calculation of \(\lambda\).

Case I:

If \(E_k<<2m_0c^2\) [ i.e. K.E is less then rest mass energy]

$$\biggl[1+\frac{E_k}{2m_0 c^2}\biggr]^{-\frac12}\approx 1$$

$$\therefore\; \lambda= \frac{h}{\sqrt{2m_0 E_k}}$$

Case II:

For an electron moving with high speed due to potential difference of V-volt.

$$\therefore \lambda=\frac{h}{\sqrt{2m_0E_k}}\biggl[1+\frac{E_k}{2m_0 c^2}\biggr]^{-\frac12}$$

$$ \lambda=\frac{h}{\sqrt{2m_0E_k}}\biggl[1+\frac{ev}{2m_0 c^2}\biggr]^{-\frac12}$$

For non-relativistic case

$$\lambda=\frac{h}{\sqrt{2m_0 ev}}$$

$$=\frac{6.626\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.602\times 10^{-19})v}}$$

$$=\frac{12.2\times 10^{-10}}{\sqrt{V}}$$

$$\therefore\;\; \lambda =\frac{12.2 A^0}{\sqrt{V}}$$

Examples:

1. Calculate the de-Broglie wavelength of electron moving with K.E 54 ev.

Solution:

Rest mass energy of electron = \(m_0\times c^2\)

$$=9.1\times 10^{-31}\times ( 3\times 10^8)^2 Joule$$

$$= \frac{9.1\times 10^{-31}\times 9\times 10^{16}}{1.602\times 10^{-19}\times 10^6} mev$$

$$\therefore\; m_0 c^2= 0.511mev$$

Given,

$$E_k= 54ev<< 2m_0 c^2$$

$$54kev<<1.022\times 10^6 ev$$

So, de-Broglie wave length is given by

$$\lambda=\frac{h}{sqrt{2m_0 ev}}=\frac{12.2A^0}{\sqrt{v}}$$

$$=\frac{12.2A^0}{\sqrt{54}}$$

$$=1.67A^0$$

\(\therefore\) de-Broglie wave length of electron moving with K.E 54 ev is \(1.67 A^0\).

2. Calculate the de-Broglie wavelength of electron moving with K.E of 1kev:

Rest mass energy of electron \(m_0 c^2\) = o.511mev

now, \(1kev<<1.022\times 10^6ev\)

So, de-Broglie wave length is given by

$$\lambda= \frac{12.2 A^0}{\sqrt{v}}=\frac{12.2A^0}{\sqrt{1000}}=0.385A^0$$

\(\therefore\) Hence de-Broglie wave length of electron having K.E 1kev is 0.385\(A^0\).

3. Calculate the de-Broglie wavelength of electron moving with total energy of 1.002mev

Solution:

Total energy (E)= 1.022mev

We have

$$E= E_k+ m_0 c^2$$

$$E_k= E- m_0 c^2$$

$$= 1.022mev- 0.511mev$$

$$=0.511mev$$

\(\therefore E_k\) is comparable to \(2m_0c^2\) we use relativistic formula.

$$\lambda= \frac{h}{\sqrt{2m_0 ev}}\biggl[1+ \frac{E_k}{2m_0 c^2}\biggr]^{-\frac12}$$

$$=\frac{12.2 A^0}{\sqrt{0.511\times 10^6}}\biggl[1+\frac{0.511mev}{1.022mev}\biggr]^{\frac{-1}{2}}$$

$$=\frac{12.2 A^0}{\sqrt{0.511\times 10^6}}\times (1+\frac12)^{\frac12}$$

\(=\)\(\dotsm\) ans

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.