Probability of electron in hydrogen nucleus for 1s and 2s wave-functions

In summary: That's not what I got.I used a different program and I got 1x10-11, but that was with using the value 5.3*10-11 for a0 that i found on google.You could try using a different program to get a more accurate result.
  • #1
guyvsdcsniper
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Homework Statement
For the 1s and 2s wave functions, find the probability of the electron to be found inside the nucleus (a sphere of radius R). A typical nucleus has a radius of 1e-15 m.
Relevant Equations
Probabily Density Function
For this problem, Is it as simple as using the probability density function, P = Ψ2 and plugging in the radius value given to me?

So essentially I am just squaring the wave function and plugging in?
 
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  • #2
quittingthecult said:
Homework Statement:: For the 1s and 2s wave functions, find the probability of the electron to be found inside the nucleus (a sphere of radius R). A typical nucleus has a radius of 1e-15 m.
Relevant Equations:: Probabily Density Function

For this problem, Is it as simple as using the probability density function, P = Ψ2 and plugging in the radius value given to me?

So essentially I am just squaring the wave function and plugging in?
Essentially yes, but plugging into what? Please show what you intend to do.
 
  • #3
So I used the radial wave functions from this website, http://plato.mercyhurst.edu/chemistry/kjircitano/InorgStudysheets/InorgWaveFunction.pdf

and basically squared each wave function. I am given the radius of a nucelus which is r=1x10-15.

So I plug that in for r in the function and also need to plug in for a.

Does that seem correct?
IMG_9361.JPG
 
  • #4
It is correct as far as squaring the wavefunctions is concerned. That's step 1. What comes next?
 
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  • #5
kuruman said:
It is correct as far as squaring the wavefunctions is concerned. That's step 1. What comes next?
It should just be plugging in. Do you know where I can find the value of a? I wasnt given that information
 
  • #6
Just leave it as a. Maybe it will cancel out. If by "plugging in" you mean replacing ##r## with ##R##, that is incorrect. You need to do an integral.
 
  • #7
It may help to review what the definition of the probability density function is and how does it relate to probability?
 
  • #8
kuruman said:
Just leave it as a. Maybe it will cancel out. If by "plugging in" you mean replacing ##r## with ##R##, that is incorrect. You need to do an integral.
So does that mean instead of integrating ∫ψ2r2dr, I should be integrating∫ψ2R2dR?
 
  • #9
quittingthecult said:
Do you know where I can find the value of a?
Actually, what you call ##a## should be ##a_0##, a.k.a. the Bohr radius. It is given in terms of other constants at the bottom of the first page of the document that you linked in post #3. You can also look it up under "Bohr radius" if you don't trust yourself to calculate its numerical value.
 
  • #10
quittingthecult said:
So does that mean instead of integrating ∫ψ2r2dr, I should be integrating∫ψ2R2dR?
I never implied that. I just didn't understand what you meant by "plug in". Integrate ##\int \psi^2(r)r^2dr##. What are the limits of integration?
 
  • #11
kuruman said:
I never implied that. I just didn't understand what you meant by "plug in". Integrate ##\int \psi^2(r)r^2dr##. What are the limits of integration?
It should be 0 to R. Sorry I misunderstood you at first but waited for you to respond. So given that the radius is typically 1x10-15m, I guess you could say the limit of integration should be 0 to 1x10-15m?
 
  • #12
quittingthecult said:
It should be 0 to R. Sorry I misunderstood you at first but waited for you to respond. So given that the radius is typically 1x10-15m, I guess you could say the limit of integration should be 0 to 1x10-15m?
Correct. Go for it.
 
  • #13
Also, don't forget the angular part of the integration.
 
  • #14
kuruman said:
Correct. Go for it.
I used wolframalpha and got this
Screen Shot 2022-04-08 at 8.35.35 PM.png


Which I guess makes sense. The probability of this happening is very unlikely, but not zero.
 
  • #15
bob012345 said:
Also, don't forget the angular part of the integration.
I thought since I was using the radial wave function I just need to integrate wrt to r^2dr?
 
  • #16
quittingthecult said:
I used wolframalpha and got this View attachment 299621

Which I guess makes sense. The probability of this happening is very unlikely, but not zero.
It would be better to do the integration symbolically first then plug in numbers. Also so we can see if it is correct.
 
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  • #17
quittingthecult said:
I thought since I was using the radial wave function I just need to integrate wrt to r^2dr?
Use the full wave function in the integration since you want the actual probability. It is true the angular part will be constant up for 1s and 2s. But it is still there.
 
  • #18
bob012345 said:
Also, don't forget the angular part of the integration.
There should be no angular part because the quoted radial wavefunction is normalized so that ##\int_0^{\infty} \psi^2(r)r^2dr=1##.
 
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  • #19
quittingthecult said:
I used wolframalpha and got this View attachment 299621

Which I guess makes sense. The probability of this happening is very unlikely, but not zero.
That's not what I got.
 
  • #20
kuruman said:
That's not what I got.
I used a different program and I got 1x10-11, but that was with using the value 5.3*10-11 for a0 that i found on google.
 
  • #21
quittingthecult said:
I used a different program and I got 1x10-11, but that was with using the value 5.3*10-11 for a0 that i found on google.

EDIT:

Now I did it by hand using the tabular method and a0 = 5.3*10-11 and, taking the absolute value, 3.72*10-32
 
  • #22
quittingthecult said:
I used a different program and I got 1x10-11, but that was with using the value 5.3*10-11 for a0 that i found on google.
That is the correct value for ##a_0##. You don't need a program for this. It is easier to integrate a polynomial so expand the exponential in Taylor series first. How many terms do you think you should keep if each one is smaller than the previous one by a factor of ##R/a_0\approx 10^{-5}?##
 
  • #23
kuruman said:
That is the correct value for ##a_0##. You don't need a program for this. It is easier to integrate a polynomial so expand the exponential in Taylor series first. How many terms do you think you should keep if each one is smaller than the previous one by a factor of ##R/a_0\approx 10^{-5}?##
Not sure if you read my edit as our post are very close but I did get a value of 3.72*10-32 after doing it by hand. Does this seem correct?

But to answer your question, I would think maybe just the first term? There isn't much change in the values if its a magnitude of 10-5
 
  • #24
quittingthecult said:
Not sure if you read my edit as our post are very close but I did get a value of 3.72*10-32 after doing it by hand. Does this seem correct?

But to answer your question, I would think maybe just the first term? There isn't much change in the values if its a magnitude of 10-5
Yes, by all means use the first term. That will give you an independent check of what the answer ought to be. The second term can only change the answer after the first 4 significant figures.
 
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  • #25
kuruman said:
Yes, by all means use the first term. That will give you an independent check of what the answer ought to be. The second term can only change the answer after the first 4 significant figures.
So the probability is nonzero. It is small but still not zero.

So the next question I have to ask is, is there any location where it is NOT possible to find an electron?

My friend is saying it can't be between energy levels. It is quantized because the only energy values are energy eigen values, so it can be in superposition of several but never between them.

I actually don't know if he is correct, he is taking QM I and I am still in modern physics so I can't really relate to what he is saying.

For me, it seems to me like the isn't a place you couldn't find an electron. I mean we just proved that the low chances of it being in the nucleus.
 
  • #26
quittingthecult said:
So the probability is nonzero. It is small but still not zero.

So the next question I have to ask is, is there any location where it is NOT possible to find an electron?

My friend is saying it can't be between energy levels. It is quantized because the only energy values are energy eigen values, so it can be in superposition of several but never between them.

I actually don't know if he is correct, he is taking QM I and I am still in modern physics so I can't really relate to what he is saying.

For me, it seems to me like the isn't a place you couldn't find an electron. I mean we just proved that the low chances of it being in the nucleus.
The problem you solved regarded the location of an electron or more accurately, the probability of it being located within a certain region wrt to the nucleus the electron is associated with. What energy level the electron occupies or the state it is in such as in the ##1s## state is not a physical location so apples and oranges but yes, it cannot be between states. So, distinguish between physical location and the state the electron is in.
 
  • #27
bob012345 said:
The problem you solved regarded the location of an electron or more accurately, the probability of it being located within a certain region wrt to the nucleus the electron is associated with. What energy level the electron occupies or the state it is in such as in the ##1s## state is not a physical location so apples and oranges.
thats fair enough, I didnt think about that.
 
  • #28
quittingthecult said:
thats fair enough, I didnt think about that.
I think you were both right in different ways.
 
  • #29
Did you do this symbolically? Say for the ##1s## wavefunction?
 
  • #30
quittingthecult said:
So the next question I have to ask is, is there any location where it is NOT possible to find an electron?

My friend is saying it can't be between energy levels. It is quantized because the only energy values are energy eigen values, so it can be in superposition of several but never between them.

I actually don't know if he is correct, he is taking QM I and I am still in modern physics so I can't really relate to what he is saying.

For me, it seems to me like the isn't a place you couldn't find an electron. I mean we just proved that the low chances of it being in the nucleus.
Your friend is wrong. The energy levels do not equate to precise radii. That's the whole point of the continuous wavefunction. If your friend imagines that an electron in the ground state must be found at the precise Bohr radius, then he is very wrong indeed.

The Bohr radius is only an "expected" or average value. The possible measurement of the electron's position follows a (continuous) probability distribution given by the modulus squared of the wavefunction.
 
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Related to Probability of electron in hydrogen nucleus for 1s and 2s wave-functions

1. What is the probability of finding an electron in the hydrogen nucleus for the 1s wave-function?

The probability of finding an electron in the hydrogen nucleus for the 1s wave-function is zero. According to the quantum mechanical model, the 1s wave-function represents the ground state of the hydrogen atom, where the electron is most likely to be found in the region surrounding the nucleus, but not inside it.

2. How does the probability of finding an electron in the hydrogen nucleus change for the 2s wave-function?

The probability of finding an electron in the hydrogen nucleus for the 2s wave-function is also zero. Similar to the 1s wave-function, the 2s wave-function represents an energy level where the electron is most likely to be found in the region surrounding the nucleus, but not inside it.

3. Why is the probability of finding an electron in the hydrogen nucleus zero for both the 1s and 2s wave-functions?

This is due to the Heisenberg uncertainty principle, which states that it is impossible to know both the exact position and momentum of a particle at the same time. The 1s and 2s wave-functions have a node at the nucleus, meaning that the probability of finding the electron at the nucleus is zero.

4. Can the probability of finding an electron in the hydrogen nucleus ever be non-zero?

No, according to the quantum mechanical model, the probability of finding an electron in the hydrogen nucleus will always be zero for the 1s and 2s wave-functions. However, for higher energy levels (e.g. 3s, 4s), the probability of finding the electron at the nucleus will increase due to the presence of additional nodes in the wave-function.

5. How does the probability of finding an electron at the nucleus relate to the size of the hydrogen atom?

The probability of finding an electron at the nucleus does not directly relate to the size of the hydrogen atom. However, as the size of the atom increases, the probability of finding the electron at the nucleus decreases due to the presence of more nodes in the wave-function. This is why the probability of finding an electron at the nucleus is higher for smaller atoms (e.g. helium) compared to larger atoms (e.g. xenon).

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