Radial probability density of hydrogen electron

[leehufford]

Homework Statement

I know this question has been asked before, but I am looking for a different kind of answer than the other poster. Bear with me here.

Problem: For a hydrogen atom in the ground state, what is the probability to find the electron between the Bohr radius a₀ and (1.01)a₀? (Hint: it is not necessary to evaluate any integrals to solve this problem.)

Homework Equations

R(r) = (2/a₀^3/2)e^(-r/a₀)

And P(r) is r² times the square of the absolute value of R(r).

The Attempt at a Solution

I'm pretty sure I know how to solve this problem using integration. However, because of the hint about not evaluating any integrals, I feel like I am expected to find the "trick" or "shortcut" to getting the probability without evaluating an integral. I was thinking about using delta r instead of r, since the interval is so small it is roughly constant over the interval.

But my final answer has the Bohr radius dropping out of the equation (which seems right- I am looking for a dimensionless probability) but my answer is about 1,000 times smaller than that of the book answer.

I was hoping I could get a push in the right direction. Thanks for reading,

Lee

 

[vela]

Show your work.

 

[leehufford]

Sorry about that.

P(r) dr = r²|(2/a₀^3/2)e^(-r/a₀)|²dr

I just realized I was substituting delta r for both dr and r, which can't be right. The probability has to depend on the distance from to nucleus, not just the difference in the distances. So my earlier method was completely, fundamentally wrong. I just don't see how to do it without integration.

I tried using the actual Bohr radius 5.29 ×10−11 m for r and 0.01 for dr but it still isn't working.

[(1.01)(5.29 ×10−11 m)]²|(4/(5.29 ×10−11 m)³)e^-2...(0.01)...

See now I run into using the Bohr radius for a₀ and r, which gives a constant exponent in the exponential function...which again probably isn't right. I am missing something here.

Thanks for the reply,

Lee

 

[leehufford]

I have also tried plugging in the two different radii separately and subtracting them, but I also got the wrong answer there. Again, I am not looking for the answer, I am looking for the concept. Thanks again,

Lee

 

[nrqed]

Sorry about that.

P(r) dr = r²|(2/a₀^3/2)e^(-r/a₀)|²dr

I just realized I was substituting delta r for both dr and r, which can't be right. The probability has to depend on the distance from to nucleus, not just the difference in the distances. So my earlier method was completely, fundamentally wrong. I just don't see how to do it without integration.

I tried using the actual Bohr radius 5.29 ×10−11 m for r and 0.01 for dr but it still isn't working.

[(1.01)(5.29 ×10−11 m)]²|(4/(5.29 ×10−11 m)³)e^-2...(0.01)...

See now I run into using the Bohr radius for a₀ and r, which gives a constant exponent in the exponential function...which again probably isn't right. I am missing something here.

Thanks for the reply,

Lee

I am not sure why you feel there is something wrong with getting a constant exponent. This is actually correct. You just have to set r = a_0 and for dr you use the difference of distances, that is 0.01 a_0 . Note that one should do an integral to get the exact result but since the difference of distance is much smaller than r, it is a good approximation to not integrate and just use for dr the difference of distances between the two points.

By the way, did you include the factor of 4 \pi coming from the integral over the angles?

 

[BvU]

Re trick: "I was thinking about using delta r instead of r, since the interval is so small it is roughly constant over the interval." Right !
$$\int_a^{a+\Delta a} P(r) dr \approx P(a) \Delta a$$

Re value: see Vela. Telepathy is not our strongest point.

[edit] apparently I didn't hit the Post reply button to end up as post #3. No matter. Re trick still applies and re vela was attended to. Helps me to ask lee to check the normalization of his P(r). (which, on closer inspection, is correct. Final answer 0.01 * 4 exp(-2) , is what I got.)

 

[vela]

Re value: see Vela. Telepathy is not our strongest point.

You read my mind!

 

[leehufford]

Thanks for all the replies, I got it now.