Hydrogen atom and probabilities

[eku_girl83]

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

 

[RedX]

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.

 

[Janitor]

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.

 

[RedX]

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

 

[Dr Transport]

If memory serves me correctly, the radial function has a factor of 1/a in it....

 

[sifeddin]

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]

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 . 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.

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! .

At your service maam

 

[jtbell]

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.)