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Chapter 8: Problem 111

In the ground state of a hydrogen atom, what is the probability of finding an electron anywhere in a sphere of radius (a) \(a_{0},\) or \(\left(\text { b) } 2 a_{0} ?\right.\)

Short Answer

Expert verified
In this problem, the aim is to calculate the probability of an electron being found in a sphere of a certain radius within a hydrogen atom's ground state. This involves applying the radial distribution function and integrating it within the desired sphere to find probabilities. Detailed answer depends on the results of the integrals.

Step by step solution

01

Recalling the wave function and the Radial Distribution Function

The wave function for ground state of hydrogen atom is given by \(\psi_{100} = \frac{1}{\sqrt{\pi a_0^{3}}}e^{-r/a_0}\), and the radial distribution function is given by \(P(r) = |\psi(r)|^2 \cdot 4 \pi r^2\). This shows the distribution of electron as function of the distance to the nucleus.
02

Probability Calculation for a Sphere of Radius \(a_{0}\)

Now, to find probability of electron being inside a sphere of radius \(a_{0}\), the radial distribution function (RDF) must be integrated from 0 to \(a_{0}\): \(\int_0^{a_{0}} P(r)\, dr\).
03

Plugging RDF into the Integral to find the Probability

We substitute the expression for RDF into the integral: \(\int_0^{a_{0}} \left(4\pi r^2 \cdot \left(\frac{1}{\sqrt{\pi a_0^{3}}}e^{-r/a_0}\right)^2\right)\, dr\). Solve this integral to get the probability of finding the electron in the sphere with radius \(a_{0}\).
04

Repeat Steps 2 and 3 for a Sphere of Radius \(2a_{0}\)

Repeat the same process, but this time change the limit of the integral to \(2a_{0}\) instead of \(a_{0}\) to get the probability of finding the electron in the sphere with radius \(2a_{0}\)

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Most popular questions from this chapter

The Pfund series of the hydrogen spectrum has as its longest wavelength component a line at \(7400 \mathrm{nm}\) Describe the electron transitions that produce this series. That is, give a Bohr quantum number that is common to this series.

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