Consider a particle of mass mand energy E in a region where the potential energy is constant U₀. Greater than E and the region extends to$\mathbf{x}\mathbf{=}\mathbf{+}\mathbf{\infty }$

(a) Guess a physically acceptable solution of the Schrodinger equation in this region and demonstrate that it is solution,

(b) The region noted in part extends from x = + 1 nm to $+\infty$. To the left of x = 1nm. The particle’s wave function is Dcos (10⁹m^-1 x). Is also greater than Ehere?

(c) The particle’s mass m is 10^-3 kg. By how much (in eV) doesthe potential energy prevailing from x=1 nm to U₀. Exceed the particle’s energy?

(a) State the expression for Schrodinger wave equation.

$\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{E\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Here, $\psi \left(x\right)$is a wave function along direction$\mathrm{U}\left(\mathrm{x}\right)$ is the potential energy of the particle, E is the particle’s energy, h is the modified Planks constant, and m is the mass of the particle.

In classical mechanics for the case E < U₀, the particle is reflected totally. That means the particle will be reflected backwards at the potential barrier. So, there is no possibility for particle to penetrate the barrier. Hence, there is no transmission of particle into the barrier.

But, in quantum mechanics for the case E < U₀, the probability corresponding to the particle’s tunnelling through the barrier is not zero. Hence, there is a possibility for the particle to penetrate the through the barrier. This phenomenon is known as the tunnelling effect.

For the condition E < U₀ , the wave traveling along positive x direction (x>0), after a small time interval, the wave will die out

The general form of exponential decaying wave function is:

$\mathrm{\psi }\left(\mathrm{x}\right)\propto {\mathrm{e}}^{-\mathrm{\alpha }\left|\mathrm{x}\right|}$

Here, $\mathrm{\psi }\left(\mathrm{x}\right)$is the wave function along x direction, $\alpha$is the wave propagation, and$\left|x\right|$ is the absolute value of position of particle along x direction.

The potential energy in the case of potential step is.

$\mathrm{U}\left(\mathrm{x}\right)=\left(\begin{array}{c}0\mathrm{forx}<0\\ {\mathrm{U}}_0\mathrm{forx}\ge 0\end{array}\right)$

Here, U₀is the potential energy of the particle in the region $\mathrm{x}⩾0$.

Consider the case E<U₀.

The wave function for the potential step for the case E<U₀ is shown below:

Here, the potential in region I is zero, and potential in region II is U₀

The Schrodinger wave equation in region II is,

$\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}{\mathbf{U}}_{\mathbf{0}}\mathbf{\psi }\mathbf{\left(}\mathbf{X}\mathbf{\right)}\mathbf{=}\mathit{E}\mathit{\psi }\mathbf{\left(}\mathbf{X}\mathbf{\right)}$

Multiply the above equation on both sides with $-\frac{2\mathrm{m}}{{\mathrm{h}}^2}$.

localid="1656118013311" $\left(-\frac{2m}{h^2}\right)\left(-\frac{h^2}{2m}\frac{d^2\psi \left(x\right)}{d{x}^2}+U_0\psi \left(X\right)\right)\mathbf{=}\left(-\frac{2m}{h^2}\right)\mathit{E}\mathit{\psi }\mathbf{\left(}\mathbf{X}\mathbf{\right)}$

localid="1656118027619" $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{2}\mathbf{m}{\mathbf{U}}_{\mathbf{0}}}{{\mathbf{h}}^{\mathbf{2}}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{2}\mathbf{m}\mathbf{E}}{{\mathbf{h}}^{\mathbf{2}}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

localid="1656118041032" $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{2}\mathbf{m}\mathbf{U}}{{\mathbf{h}}^{\mathbf{2}}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\frac{\mathbf{2}\mathbf{m}\mathbf{E}}{{\mathbf{h}}^{\mathbf{2}}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{0}$

localid="1656118055211" $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{2}\mathbf{m}{\mathbf{U}}_{\mathbf{0}}}{{\mathbf{h}}^{\mathbf{2}}}\left(U_0-E\right)\mathbf{\psi }\left(X\right)\mathbf{=}\mathbf{0}$

Let, localid="1656118070285" ${\mathbf{a}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}\mathbf{m}\left(U_0-E\right)}{{\mathbf{h}}^{\mathbf{2}}}$

Substitute a² for$\frac{2\mathrm{m}}{{\mathrm{h}}^2}\left({\mathrm{U}}_0-\mathrm{E}\right)$in$\frac{{\mathrm{d}}^2\mathrm{\psi }\left(\mathrm{x}\right)}{{\mathrm{dx}}^2}-\frac{2\mathrm{m}}{{\mathrm{h}}^2}\left({\mathrm{U}}_0-\mathrm{E}\right)\mathrm{\psi }\left(\mathrm{X}\right)=0$

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\psi }\left(x\right)}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{-}{\mathbf{a}}^{\mathbf{2}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{0}$………………………. (1)

The general solution of the above wave function is of the following form,

$\mathbf{\psi }\left(x\right)\mathbf{=}{\mathbf{Ee}}^{\mathbf{ax}}\mathbf{+}{\mathbf{Ce}}^{\mathbf{-}\mathbf{ax}}$……………………….. (2)

Here, E, and C are the constants.

From the postulates of wave function, the wave function must be finite everywhere.

Let us verify whether the wave function is valid or not.

When$x\to \infty$, the termis e^ax diverges. That is $e^{a(\infty )}=0$. Here, the wave function is not infinite.

So, in equation (2) the term ${\mathrm{Ee}}^{\mathrm{ax}}$is not acceptable. This is possible as the constant E is zero.

When $\mathrm{x}\to \infty$, the term e^-ax is the coverage. That is. ${\mathrm{e}}^{-\mathrm{a}\left(\infty \right)}=0$.here, the wave function is finite.

So, in equation (2) the term Ce^-ax is acceptable.

Substitute 0 for E in $\mathrm{\psi }\left(\mathrm{x}\right)={\mathrm{Ee}}^{\mathrm{ax}}+{\mathrm{Ce}}^{-\mathrm{ax}}$.

$\mathrm{\psi }\left(\mathrm{x}\right)=\left(0\right){\mathrm{e}}^{\mathrm{ax}}+{\mathrm{Ce}}^{-\mathrm{ax}}$

$={\mathrm{Ce}}^{-\mathrm{ax}}$

Therefore, the solution of the Schrodinger wave equation in region II is $\mathrm{\psi }\left(\mathrm{x}\right)={\mathrm{Ce}}^{-\mathrm{ax}}$.

(b) The wave function for the left of x = 1 nm is,

$\mathrm{\psi }{\left(\mathrm{x}\right)}_{0⩽\mathrm{x}<1\mathrm{nm}}=\mathrm{Dcos}\left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right)$

For the region $0⩽\mathrm{x}<1$nm, the potential energy is constant which is U₀ . And given that in region U₀>E.

For the region $0⩽\mathrm{x}<1$nm, the potential energy is U(x). The particle’s wave function for region $0⩽x<1$nm, is $\mathrm{Dcos}\left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right)$. The particle is localized in this region; this is possible as the potential energy is must less than the particle’s total energy.

Therefore, the answer is No. That is, the potential energy is not greater than the particle’s energyE.

(c) The wave function for region$0⩽\mathrm{x}<1$nm is,

$\mathrm{\psi }{\left(\mathrm{x}\right)}_{0⩽\mathrm{x}<1\mathrm{nm}}=\mathrm{Dcos}\left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right)$

The first derivate of wave function is:

$\frac{\mathrm{d\psi }{\left(\mathrm{x}\right)}_{0⩽\mathrm{x}<1\mathrm{nm}}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{Dcos}\right({10}^9{\mathrm{m}}^{-1}\mathrm{x}\left)\right)$

$=-\left({10}^9{\mathrm{m}}^{-1}\right)\mathrm{Dsin}\left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right)$

The wave function for region$\mathrm{x}⩾1\mathrm{nm}$is,

$\mathrm{\psi }{\left(\mathrm{x}\right)}_{\mathrm{x}⩾1\mathrm{nm}}={\mathrm{Ce}}^{-\mathrm{ax}}$

So, the derivate is:

$\frac{\mathrm{d\psi }{\left(\mathrm{x}\right)}_{\mathrm{x}⩾1\mathrm{nm}}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{Ce}}^{-\mathrm{ax}}\right)$

$=-{\mathrm{aCe}}^{-\mathrm{ax}}$

From the properties of wave function, apply the boundary condition for first derivative of $\mathrm{\psi }\left(\mathrm{x}\right)$at x = 1 nm

$\frac{\mathbf{d\psi }{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}_{\mathbf{0}\mathbf{⩽}\mathbf{x}\mathbf{<}\mathbf{1}\mathbf{nm}}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{d\psi }{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}_{\mathbf{x}\mathbf{⩾}\mathbf{1}\mathbf{nm}}}{\mathbf{dx}}$

Substitute $-\left({10}^9{\mathrm{m}}^{-1}\right)\mathrm{Dsin}\left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right)$for $\frac{\mathrm{d\psi }{\left(\mathrm{x}\right)}_{0⩽\mathrm{x}<1\mathrm{nm}}}{\mathrm{dx}}$ and, -aCe^-ax for $\frac{\mathrm{d\psi }{\left(\mathrm{x}\right)}_{\mathrm{x}⩾1\mathrm{nm}}}{\mathrm{dx}}$

$-\left({10}^9{\mathrm{m}}^{-1}\right)\mathrm{Dsin}\left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right)=-{\mathrm{aCe}}^{-\mathrm{ax}}$

$\mathrm{a}=\left({10}^9{\mathrm{m}}^{-1}\right)\frac{\sin \left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right)}{\cos \left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right)}$

$=\left({10}^9{\mathrm{m}}^{-1}\right)\tan \left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right)$

Substitute 1nm for x in$\mathrm{a}=\left({10}^9{\mathrm{m}}^{-1}\right)\tan \left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right)$

$\mathrm{a}=\left({10}^9{\mathrm{m}}^{-1}\right)\tan \left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\right(1\mathrm{nm}\left)\right)$

$=\left({10}^9{\mathrm{m}}^{-1}\right)\tan \left({10}^9{\mathrm{m}}^{-1}\mathrm{x}\left(1\mathrm{nm}\right)\right)\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)$

$=\left({10}^9{\mathrm{m}}^{-1}\right)\tan \left(1\right)$

$=\left({10}^9{\mathrm{m}}^{-1}\right)(1.5574)$

Substitute $\left({10}^9{\mathrm{m}}^{-1}\right)(1.5574)$ for a, and 10^-30 kg for m in${\mathrm{U}}_0-\mathrm{E}=\frac{{\mathrm{a}}^2{\mathrm{h}}^2}{2\mathrm{m}}$

${\mathrm{U}}_0-\mathrm{E}=\frac{{\left(\left({10}^9{\mathrm{m}}^{-1}\right)\left(1.5574\right)\right)}^2{\left(1.05457\times {10}^{-34}\mathrm{J}\right)}^2}{2\left({10}^{-30}\mathrm{kg}\right)}$

$=1.3487\times {10}^{-20}\mathrm{J}$

Rounding off to three significant figures, the potential energy exceeds the particle’s energy by 1.35 x 10^-20 J.

Convert the value of ${\mathrm{U}}_0-\mathrm{E}$from J to eV as follow:

${\mathrm{U}}_0-\mathrm{E}=1.3487\times {10}^{-20}\mathrm{J}$

$=\left(1.3487\times {10}^{-20}\mathrm{J}\right)\left(\frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right)$

= 0.08429eV

Rounding off to three significant figures, the potential energy exceeds the particle’s energy by = 0.08429eV.