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Bohr radius 
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#1
Apr212, 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 1D 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? 


#2
Apr312, 04:57 AM

P: 281

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. 


#3
Apr312, 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? 


#4
Apr312, 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.



#5
Apr312, 11:52 AM

P: 164

So you just equate the coefficients and then solve for both b and E, right?



#6
Apr312, 03:07 PM

P: 281

Yeah.



#7
Apr412, 11:33 AM

P: 164

Thanks.



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