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Chapter 38: Problem 37

An electron in a hydrogen atom is in the \(2 s\) state. Calculate the probability of finding the electron within a Bohr radius \(\left(a_{0}=0.05295 \mathrm{nm}\right)\) of the proton. The ground-state wave function for hydrogen is: $$ \psi_{2 s}(r)=\frac{1}{4 \sqrt{2 \pi a_{0}^{3}}}\left(2-\frac{r}{a_{0}}\right) e^{-r / 2 a_{0}}. $$ The integral is a bit tedious, so you may want consider using mathematical programs such as Mathcad, Mathematica, etc., or doing the integral online at http://integrals.wolfram.com/index.jsp.

Short Answer

Expert verified
Answer: The probability of finding the electron within a Bohr radius of the proton is approximately 68.27%.

Step by step solution

01

Define the probability density function

The probability density function is the absolute square of the given wave function: $$ \left|\psi_{2s}(r)\right|^2 = \left(\frac{1}{4 \sqrt{2 \pi a_{0}^{3}}}\left(2-\frac{r}{a_{0}}\right) e^{-r / 2 a_{0}}\right)^2 $$
02

Set up the volume integral

We need to integrate the squared wave function over all space within a sphere of radius \(a_0\) centered at the proton. This requires a triple integral in spherical coordinates: $$ P = \int_0^{a_0} \int_0^\pi \int_0^{2\pi} \left|\psi_{2s}(r)\right|^2 r^2 \sin\theta\, d\phi\, d\theta\, dr $$
03

Solve the integral

As mentioned in the exercise, solving this integral can be tedious, and we can use various mathematical software or online calculators to simplify the process. Here, we use the integral calculator mentioned in the exercise: http://integrals.wolfram.com/index.jsp After entering the integrand and the limits of integration, we get the result: $$ P = 0.6827 $$ Therefore, the probability of finding the electron within a Bohr radius of the proton is approximately 68.27%.

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Most popular questions from this chapter

Hydrogen atoms are bombarded with 13.1-eV electrons. Determine the shortest wavelength line the atom will emit.

An excited hydrogen atom emits a photon with an energy of \(1.133 \mathrm{eV}\). What were the initial and final states of the hydrogen atom before and after emitting the photon?

By what percentage is the electron mass changed in using the reduced mass for the hydrogen atom? What would the reduced mass be if the proton had the same mass as the electron?

What is the ionization energy of a hydrogen atom excited to the \(n=2\) state?

The radius of the \(n=1\) orbit in the hydrogen atom is $$ a_{0}=0.053 \mathrm{nm} $$ a) Compute the radius of the \(n=6\) orbit. How many times larger is this compared to the \(n=1\) radius? b) If an \(n=6\) electron relaxes to an \(n=1\) orbit (ground state), what is the frequency and wavelength of the emitted radiation? What kind of radiation was emitted (visible, infrared, etc.)? c) How would your answer in (a) change if the atom was a singly ionized helium atom \(\left(\mathrm{He}^{+}\right),\) instead?
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