The probability of an electron being found in a certain region - probably easy?

In summary, the conversation discusses the hydrogen ground state wavefunction, which is represented by \psi(r,\theta,\phi) = (\frac{1}{a^3^/^2\sqrt{\pi}})e^-^r^/^a. The probability of finding the electron within the nucleus is calculated using the identity \int x^2e^-^xdx = b^3/3, and the equation P(r) = \int |\psi|^2 dV = \int |\psi|^2 4\pi r^2dr is used. However, a wrong substitution is initially made, resulting in incorrect units of m3. A new substitution of y=\frac{2r}{a} is suggested to solve
  • #1
jeebs
325
4
the hydrogen ground state has the following wavefunction

[tex]\psi[/tex](r,[tex]\theta[/tex],[tex]\phi[/tex]) = [tex](\frac{1}{a^3^/^2\sqrt{\pi}})e^-^r^/^a[/tex]

where a is the bohr radius a = 5.29x10-11m
I have to calculate the probability of finding the electron within the nucleus (radius R = 10-15m), and I am given an identity:

[tex]\int x^2e^-^xdx = b^3/3[/tex] when this integral is evaluated between 0 and b.


I know that the probability is given by P(r) = [tex]\int |\psi|^2 dV[/tex] = [tex]\int |\psi|^2 4\pi r^2dr[/tex] evaluating between 0 and R.

so when I put [tex]\psi[/tex] into the probability equation, I get

P(r) = [tex](4/a^3)\int r^2 e^-^2^r^/^a dr[/tex] and I then try to use that identity, by making the substitution that x=kr (where k = 2/a)
so that dx = kdr and x2 = (kr)2

which gives me [tex]k^3\int r^2 e^-^k^r dr = R^3/3[/tex]

which yields P(r) = [tex](4/a^3)\int r^2 e^-^k^r dr = (4/a^3)R^3/3k^3 [/tex] which has units of m3.

clearly I am going wrong because I am getting units of volume. this is frustrating me now, where am I going wrong?

thanks.
 
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  • #2
i think u used the wrong substitution...
your integral isn't of the same form as that given in the hint.

try [itex]y=\frac{2r}{a}[/itex]. hopefully that will help.
 
  • #3
even with your substitution this still gives me units of m3...
 

1. What is the probability of finding an electron in a specific region?

The probability of finding an electron in a specific region is determined by the wave function, also known as the probability density function. This function describes the likelihood of finding an electron at a particular point in space.

2. How is the probability of finding an electron calculated?

The probability of finding an electron is calculated by integrating the square of the wave function over the region of interest. This gives the probability of finding the electron within that region.

3. Is the probability of finding an electron constant throughout a region?

No, the probability of finding an electron can vary within a region. This is due to the wave nature of electrons and the uncertainty principle, which states that it is impossible to know the exact position and momentum of an electron at the same time.

4. Can the probability of finding an electron be greater than 1?

No, the probability of finding an electron cannot be greater than 1. This would imply that the electron is guaranteed to be found within the region, which is not possible according to the uncertainty principle.

5. How does the probability of finding an electron change with distance from the nucleus?

The probability of finding an electron decreases as the distance from the nucleus increases. This is due to the shape of the wave function, which has a higher amplitude closer to the nucleus and a lower amplitude further away.

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