Bohr radius


by glebovg
Tags: bohr, radius
glebovg
glebovg is offline
#1
Apr2-12, 04:23 PM
P: 164
1. The problem statement, all variables and given/known data

By substituting the wave function [itex]\psi (x) = Ax{e^{ - bx}}[/itex] into the Schoedinger equation for a 1-D atom, show that a solution can be obtained for [itex]b = 1/{a_0}[/itex], where [itex]{a_0}[/itex] is the Bohr radius.

2. Relevant equations

[itex] - \frac{{{\hbar ^2}}}{{2m}}\frac{{{d^2}\psi (x)}}{{d{x^2}}} - \frac{{{e^2}}}{{4\pi {\varepsilon _0}x}}\psi (x) = E\psi (x)[/itex].

[itex]{a_0} = \frac{{4\pi {\varepsilon _0}{\hbar ^2}}}{{m{e^2}}}[/itex]

3. The attempt at a solution

I get to the point where [itex]2b - x{b^2} = \frac{{2m}}{{{\hbar ^2}}}Ex + \frac{m}{{{\hbar ^2}}}\frac{{{e^2}}}{{2\pi {\varepsilon _0}}}[/itex]. If I let x = 0 I get the desired result. Can I do that?

Apparently the wave function solving the equation must satisfy two conditions:

[itex]\psi (x) \to 0[/itex] as [itex]x \to \infty[/itex]
[itex]\psi (0) = 0[/itex]

But why is that? Can anyone explain?
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JesseC
JesseC is offline
#2
Apr3-12, 04:57 AM
P: 281
Quote Quote by glebovg View Post

[itex]\psi (x) \to 0[/itex] as [itex]x \to \infty[/itex]
[itex]\psi (0) = 0[/itex]

But why is that? Can anyone explain?
The first condition is necessary because the wavefunction must be normalisable. Meaning it must have a finite integral over all space. If it tended to anything other than 0, then this would not be the case.

The second condition is necessary because you are talking about an atom. Atoms have a positive nucleus. There is no probability of finding an electron at exactly the centre of the nucleus.
glebovg
glebovg is offline
#3
Apr3-12, 08:08 AM
P: 164
Can you help? How do I show that [itex]b = 1/{a_0}[/itex]?

What should E be?

Also, how to derive the expression for the ground state energy?

JesseC
JesseC is offline
#4
Apr3-12, 09:23 AM
P: 281

Bohr radius


You need to recognise that in the ground state [itex]x = a_0 = 1/b[/itex] and therefore [itex]E = -\frac{b^2 \hbar^2}{2m} [/itex]. Thus you can cancel the terms for [itex]x[/itex] from the equation and calculate the bohr radius by plugging in known values of the constants. This also leaves you with an equation for calculating the ground state energy.
glebovg
glebovg is offline
#5
Apr3-12, 11:52 AM
P: 164
So you just equate the coefficients and then solve for both b and E, right?
JesseC
JesseC is offline
#6
Apr3-12, 03:07 PM
P: 281
Yeah.
glebovg
glebovg is offline
#7
Apr4-12, 11:33 AM
P: 164
Thanks.


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