Penetration depth and potential barrier.

Penetration Depth ( \(\Delta x\)):

It is the distance from x=o at which the value of wave function becomes 1/e times its value at x=0. It is denoted by (\(\Delta x\)) and can be determined as

$$\psi_{II}(x= \Delta x)= \frac{1}{e} x \psi_{II}(x=0)$$

$$or,\; Ce^{-k_2\cdot \Delta x}= \frac{1}{e} ce^{-k_2\cdot 0}$$

$$or,\;\; e^{-k_2\Delta x}= \frac{1}{e}$$

$$or,\;\; e^{-k_2\Delta x}= e^{-1}$$

$$or, -k_2\Delta x= -1$$

$$or,\; \Delta x= \frac{1}{k_2}\dotsm(19)$$

From equation (8)

$$K_2=\sqrt{\frac{2m(V_0-E)}{\hbar^2}}$$

$$\therefore\; \Delta x= \frac{\hbar}{\sqrt{2m(V_0-E)}}\dotsm(20)$$

$$\therefore\; \; \Delta x\propto \frac{1}{\sqrt{m}}$$

$$\propto \frac{1}{\sqrt{V_0-E}}$$

Special case: For \(V_0=\infty\) ( infinity height of step):-

$$\Delta x=0 \; \psi_{II}(x)= ce^{-k_2 x}= ce^{-\infty} = 0$$

For highly massive particle:

$$\Delta x\approx 0\; (small)$$

In Quantum Mechanics there is 100% chance the particle is reflected but through different distance from x=0.

Potential Barrier { Rectangular or squre potential barrier ] [\( E> V_0\)] [ Case I] :

A potential step having finite width is known as potential barrier.

It is defined as,

$$V(x)= V_0\; For \; 0\leq x\leq a$$

$$=0; otherwise \dotsm(1)$$

consider, a flux of particle having some energy 'E' and mass 'm' be incident on the potential barrier of the form given by equation (1) from left.

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the wave mechanical behaviour of particle in different regions can be obtained by solving one dimensional independent Schrodinger equation.

$$\frac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+ V(x)\psi(x)= E\psi(x)$$

$$or,\; \frac{d^2\psi(x)}{dx^2}+ \frac{2m}{\hbar^2}[E-V(x)]\psi(x)=0\dotsm(2)$$

In region (I) x<0,

$$\psi(x)= \psi_I(x)$$

$$V(x)=0$$

Then equation (2) becomes

$$or,\; \frac{d^2\psi_I(x)}{dx^2}+ \frac{2mE}{\hbar^2}\psi_I(x)=0$$

$$or,\; \frac{d^2\psi_I(x)}{dx^2}+K_1^2 \psi_I(x)=0\dotsm(3)$$

Where, \(K_1= \sqrt{\frac{2mE}{\hbar^2}}\)

and solution of equation (3) is,

$$\psi_I(x)= Ae^{ik_1x}+ Be^{-k_1 x}\dotsm(4)$$

In region (II) \(0\leq x\leq a ,\psi(x)= \psi_{II}(x), V(x)= V_0<E\) equation (2) becomes,

$$\frac{d^2\psi_{II}(x)}{dx^2}+ \frac{2m}{\hbar^2}(E-V_0)\psi_{II} (x)=0$$

$$or,\; \frac{d^2\psi_{II}(x)}{dx^2}+ K_2^2 \psi_{II}(x)=\dotsm(5)$$

Where, \(K_2= \sqrt{\frac{2m(E-V_0)}{\hbar^2}}\)

Solution of (5) is $$\psi_{II}(x)= ce^{ik_2 x}+ De^{-ik_2 x} \dotsm(6)$$

Where, \(ce^{ik_2 x}\) represents transmitted wave in region (2) from x=0 and \(De^{-ik_2 x}\) represents reflected wave from x=0.

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.