Hydrogen atom and probabilities

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Here's my question...
For a hydrogen atom in the ground state, what is the probability to find the electron between 1.00a and 1.01a, where a is the Bohr radius? It is not necessary to evaluate any integrals to solve this problem.

I know that P(r)=r^2*(R(r))^2. I used the R(r) expression for n=1, l=0 and then substituted values r=1a and r=1.01a
I subtracted P(1a)-P(1.01a) to get .0000537/a
Expressed as a percent, this is .0054/a
The answer in the back of the book is .0054
Why is my answer off by a factor of 1/a?

Any help would be appreciated!
Thanks
 

Answers and Replies

  • #2
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P is a probability distribution, so you would take like P(1a) and multiply is by (1a)^2 and then multiply it by dr, which would be (.01)a. And then to get a percent you'd multiply by 100.
 
  • #3
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Don't know if this helps-

It seems to me that your volume element, given that l=0 implies spherical symmetry, is 4 pi r^2 delta r (an approximation that becomes exact in the limit where delta r --> 0).

Probability has no physical units/dimensions, whereas the Bohr radius a obviously does have units of length, so book's answer looks plausible in that sense.
 
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  • #4
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The spherical harmonics which accompany the radial part are usually normalized. So r^2 R(r)^2 dr is usually normalized too. So basically I don't think you need a 4pi or whatever. Probability densities do have units. r^2dr has units of volume, and R(r)^2 should have units of 1 over volume. I mean say that the probablity density of occupying the Bohr radius is .5, which seems reasonable.

Then:

r^2 R(r)^2 dr =(1a)^2 * (.5/a^3) * (.01a)=.005
 
  • #5
Dr Transport
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If memory serves me correctly, the radial function has a factor of 1/a in it.............
 
  • #6
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eku_girl83 said:
Here's my question...
For a hydrogen atom in the ground state, what is the probability to find the electron between 1.00a and 1.01a, where a is the Bohr radius? It is not necessary to evaluate any integrals to solve this problem.
I guess you are wrong! It is necessary to evaluate the probability integral to solve this problem, i.e. integrate abs(psi_r)^2 over r from 1.00a to 1.01a [psi_r is the wavefunction in per r dimension in the r variable for sake of clarity]

eku_girl83 said:
I know that P(r)=r^2*(R(r))^2. I used the R(r) expression for n=1, l=0 and then substituted values r=1a and r=1.01a
I subtracted P(1a)-P(1.01a) to get .0000537/a
Expressed as a percent, this is .0054/a
I hope that P(r) you know is the cumulative probability function with r otherwise you are mis-functioning :wink: . If it is so then it must be dimensionless when you substitute with radial distances using the same units you originally got R(r) and integrated it on. The percent of course is right.

eku_girl83 said:
The answer in the back of the book is .0054
Why is my answer off by a factor of 1/a?

Any help would be appreciated!
Thanks
I think you need to re-work the problem in the light of my and others directions.
After all don't you think it's too small even for ratio! :surprised .

At your service maam
 
  • #7
jtbell
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sifeddin said:
I guess you are wrong! It is necessary to evaluate the probability integral to solve this problem, i.e. integrate abs(psi_r)^2 over r from 1.00a to 1.01a [psi_r is the wavefunction in per r dimension in the r variable for sake of clarity]
This is true if you want an exact answer. However, if the probabilty density doesn't change significantly over the interval from r=1.00a to r=1.01a, you can get a good approximation by evaluating the probability density at some point in that interval, say r=1.00a for simplicity, and multiplying it by the volume of a thin spherical shell, which is approximately 4*pi*r^2 * thickness. The radius of the shell is a and the thickness is 0.01a.

(This assumes that the wave function is normalized to begin with.)
 
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