Probability of electron location in Hydrogen atom

[leroyjenkens]

Homework Statement

What is the probability that an electron in the ground state of hydrogen is within one Bohr radius of the nucleus?

Homework Equations

$$P_{nl}(r) = r^{2}|R_{nl}(r)|^{2}$$

The Attempt at a Solution

Since it's an electron in the ground state of a hydrogen atom, that means n = 1, and that means it's in the s orbital, which means l = 0.

So using the formula provided in the book for $R_{10}(r)$, which is $\frac{2}{(a_{0})^{\frac{3}{2}}}e^{\frac{-r}{a_{0}}}$

I just square that whole thing and get $\frac{4e^{\frac{-2r}{a_{0}}}}{(a_{0})^{3}}$

I know the value of $a_{0}$, but I'm not sure what r is. Is r the radius, which happens to be the same as the Bohr radius ($a_{0}$) for this problem?
I want to be able to calculate an actual number instead of having an answer with variables in it.

Thanks.

 

[bmb2009]

The probability is the integral of the quantity you squared. The bounds the integral should be 0 to the bohr radius... that will give you the probability of the electron being within the bohr radius. Does this make sense? Check your formula for the probability... seems as if you left off the integral entirely.

 

[leroyjenkens]

Thanks. Yeah that makes sense. The book shows it with a dr on both sides without the integral sign. But yeah, the integral sign should be there.

Now all I need to do is figure out how to properly do that integral. That's just a matter of time.

Thanks again.

 

[bmb2009]

Remember you can check your answer (or compute it entirely if your instructor let's you) with wolframalpha

 

[paranormal]

I can provide a response to the question by explaining the concept of electron probability distribution in an atom. In quantum mechanics, the location of an electron cannot be determined precisely, but it can be described by a probability distribution function. This function, denoted by P(r), gives the probability of finding an electron at a distance r from the nucleus.

In the case of a hydrogen atom, the electron probability distribution function, P(r), for the ground state (n=1) can be calculated using the formula P_{10}(r) = r^{2}|R_{10}(r)|^{2}. This formula takes into account the radial probability density function, R_{10}(r), which depends on the principal quantum number (n) and the distance from the nucleus (r).

For the ground state of hydrogen, n=1 and l=0, which means the electron is in the s orbital. Plugging these values into the formula, we get P_{10}(r) = \frac{4e^{\frac{-2r}{a_{0}}}}{(a_{0})^{3}}. This function gives the probability of finding the electron at any distance r from the nucleus.

To answer the question, we need to find the probability of finding the electron within one Bohr radius (a_{0}) of the nucleus. This means we need to evaluate the probability distribution function at r=a_{0}. Substituting this value into the function, we get P_{10}(a_{0}) = \frac{4e^{\frac{-2a_{0}}{a_{0}}}}{(a_{0})^{3}} = \frac{4}{(a_{0})^{3}}.

Using the value of the Bohr radius, a_{0} = 0.529 \times 10^{-10} m, we can calculate the probability as P_{10}(a_{0}) = \frac{4}{(0.529 \times 10^{-10} m)^{3}} = 0.020. This means that there is a 2% chance of finding the electron within one Bohr radius of the nucleus in the ground state of hydrogen.

In conclusion, the probability of finding an electron in the ground state of hydrogen within one Bohr radius of the nucleus is 2%. This probability increases with higher energy levels and decreases as the distance from the nucleus increases.