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Why can't there be any continuous energy band in an atom?

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Let $E = KE +U$ be the total energy . We know the momentum operator & total-energy operator are $$\hat{p} = \frac{\hbar}{i} \frac{\partial}{\partial x} \\\\\\\\\\\\ \hat{E} = i\hbar\dfrac{\partial}{\partial t}$$ . This prompts us to write $$\hat{E} = \hat{KE} + \hat{U} \implies \hat{E} = \dfrac{\hat{p}^2}{2m} + U \implies \hat{E} = i\hbar\dfrac{\partial}{\partial t} = - \dfrac{\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2} +U$$ This special operator is the Hamiltonian operator $\hat{H}$.

Steady-state Schrodinger's equation is given by: $$\dfrac{\partial^2 \psi}{\partial x^2} + \dfrac{2m}{\hbar^2} (E - U)\psi = 0$$. In order to solve the equation, "boundary conditions" will have to be satisfy which restrict the values of $E$ which only do satisfy the equation. This is more evident in this form of equation $\hat{H} \psi_n = E_n \psi_n$ . For every hard $\psi$, there must be a corresponding eigenvalues of energy. Thus there are only certain energy states which is valid for certain "boundary conditions$^1$".


$^1$ Let you take the famous particle in a box problem.

Here the boundary conditions are that the particle's probability to find outside the box is $0$ i.e. $\psi = 0$ for $x \leq 0 \quad \& \quad x \geq L$; $L$ being length of the box. . When you solve the equation using this condition $$\psi = A\sin\dfrac{\sqrt{2mE}}{\hbar} x$$; using the condition it is infered that $\dfrac{\sqrt{2mE}}{\hbar} L = n\pi$ which eventually provides the energy $E = \dfrac{n^2\pi^2 \hbar^2}{2ml^2} \qquad n=1,2,3\cdots$. Thus, you can see energy can take only certain values owing to satisfy the "boundary conditions."

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An electron in an atom is in a bound state. Since it is a bounded particle, can analyse the problem with particle in a one dimensional potential box. Consider a one-dimensional crystal lattice with lattice constant $L$. We assume that the particle is free to move within this distance and cannot go outside. So we have two potential barriers that tend to infinity at the points $x=0$ and $x=L$ as shown in the figure. This infinite potential is to mention that the electron is well confined in that region and cannot go outside.

enter image description here

Inside the region, the potential energy is zero and outside it is infinite. So, we can define the potential as

$$V(x)=0\space for\space 0<x<L;\space V(x)=\infty \space for\space x<0, \space x>L$$

Now, we define the wave function of an electron as a function of distance $\psi(x)$. Then, the Schrodinger's equation can be written as

$$\frac{d^2 \psi(x)}{dx^2}+\frac{2mE}{\bar h^2}\psi(x)=0$$

We assume the general solution be of the form :

$$\psi(x)=A\sin kx+B\cos kx$$

where $A$ and $B$ are arbitrary constants, to be determined from the boundary conditions.

Now, we are going to apply the boundary conditions:

At $x=0$, the potential barrier is infinite. Hence the wave function vanishes at $x=0$ since there is a zero probability of finding the electron. Also the electron cannot be present at $x=L$ and hence there also the wave function vanishes. We have already mentioned that the electron is allowed to move between the lattice points only. So the wave function is localized at the region 0

$$\psi(x=0)=0; \space \psi(x=L)=0$$

The first boundary condition gives us the constant $B=0$. So, we need to find the value of $A$ only. The second boundary condition yields $A\sin kL=0$. But $A\neq 0$ since this will yield us nothing. Hence

$$\sin kL=0$$

which gives $\displaystyle{kL=n\pi\Rightarrow k=\frac{n\pi}{L}}$

where $n=1,2,3,\cdots$. $n\neq 0$ since $n=0$ ($k=0$) means the wave function $\psi(x)=0$ everywhere within the box and we came all the way for nothing. So $n$ can have only positive integers.

Also, it is to be noted that the propagation constant is having integral values corresponding to different values of $n$. Hence the wave function also have integral values given by

$$\psi_n(x)=A\sin \frac{n\pi x}{L};\space\space\space 0<x<L$$.

Our electron is well confined within the region $0<x<a$. Hence the probability of finding an electron in this region is 1. This is called normalizing the wave function, which will give us the value of $A$.

$$\int_0^a |\psi_n|^2 dx=1$$

or $$A^2\int_0^a sin^2 \left(\frac{n\pi x}{L} \right) dx=1$$.

On integration, we get

$$A=\sqrt{\frac{2}{L}}$$

So the normalized wave function becomes

$${\color{green}{\psi_n(x)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}} };\space\space\space 0<x<L$$.

and $n=1,2,3,\cdots$

Now, using this value and substituting it in the Schrodinger equation yields us the value of the energy of the electron corresponding to the wave function $\psi_n(x)$.

Thus, $${\color{blue}{E_n=\frac{h^2n^2}{8mL^2}} } ;\space \space \space n=1,2,3,\cdots$$

Thus, we have the electron energy levels are quantized. So the energy levels are discrete, not continuous as expected from a classical point of view.

The lowest energy level possible is

$$E_1=\frac{h^2}{8mL^2}$$.

From the above equation of energy it is clear that

$$E_n=n^2E_1$$

This means the spacing between two consecutive energy levels increases as

$$(n+1)^2E_1-n^2E_1=(2n+1)E_1$$

Thus the energy level diagram looks like this

enter image description here

So, as classical mechanics says, there can't be continuous range of energy levels or bands. However, if the particle becomes heavier and the length of the crystal is very large, the energy levels will be spaced very close together and eventually may become continuous. for example, if $L=1cm$, then

$$E_n-E_{(n\pm 1)}\sim 3.5\times 10^{-19} eV$$

The energy spectrum for such cases seems practically continuous. Thus the wave equation predicts that the bound particles (electrons) are associated with a discrete energy spectrum and the free particles with a continuous spectrum.

An electron is a quantum mechanical particle and hence it must obey quantum mechanical principes. Discreteness is one speciality of quantum mechanics which gradually appears as continuum as we see in our everyday life. This is the essence of Bohr's correspondence principle which states that the quantum mchanics gradually reduces to claasical mechanics at the limit of large quantum numbers.

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Why can't there be any continuous energy band in an atom?

This is the basic reason quantum mechanics had to be invented.

Once the existence of positive and negative charges was discovered, Maxwell's equations when solved for a planetary model of a central positive charge and an orbiting negative one, are completely unstable, in contrast to the gravitational problem. This is due to the fact that accelerated charges radiate electromagnetic radiation and lose energy. An orbiting electron in the field of a nucleus will continually radiate energy and fall on the nucleus as far as a classical model goes.

This means that atoms would be a soup of positive and negative charges and when an electron fell on an ion, a continuous spectrum would be observed.

This is not what was observed. Hydrogen atoms had a distinct electromagnetic spectrum, made up of lines that could be mathematically fitted very well by the Balmer and Lyman series.

balmer

That is why the Bohr model, a planetary model of the hydrogen atom was proposed, which forced stable, quantized, orbits and could derive the series appearing experimentally.

After that came the Schrodinger equation and quantum mechanics, which gave a theoretical model that describes nature in the microcosm, and is the underlying level of all classical theories.

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  • $\begingroup$ Minor point: "This is the basic reason quantum mechanics had to be invented" or that classical mechanics was inadequate and quantum mechanics was discovered? $\endgroup$ – jim May 13 '16 at 8:18

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