6–20. Calculate the probability that a hydrogen 1s electron will be found within a distance 2a, from the nucleus.
6–20. Calculate the probability that a hydrogen 1s electron will be found within a distance 2a, from the nucleus.
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This is a problem from the book with the answers. Can you please show step by step for 6-20 and 6-21. Please
![**6-20. Probability Calculation for Hydrogen 1s Electron**
Calculate the probability that a hydrogen 1s electron will be found within a distance \(2a_0\) from the nucleus.
---
This problem is similar to Example 6-10. The wave function for the 1s orbital of hydrogen is:
\[
\psi_{100} = \frac{1}{\sqrt{\pi}} \left( \frac{1}{a_0} \right)^{3/2} e^{-\sigma}
\]
where \(\sigma = r/a_0\), and the probability that the electron will be found within a distance \(2a_0\) from the nucleus is
\[
\begin{align*}
\text{prob} &= \int_0^{2\pi} d\phi \int_0^{\pi} \sin \theta \, d\theta \int_0^{2a_0} dr \, r \left( \frac{1}{\pi} \left( \frac{1}{a_0} \right)^3 \right) e^{-2\sigma} \\
&= 4 \left( \frac{1}{a_0} \right)^{3} \int_0^{2a_0} dr \, r^2 e^{-2\sigma} = 4 \int_0^{2} d\sigma \, \sigma^2 e^{-2\sigma} \\
&= 4 \left( \frac{1}{4} - \frac{13}{4} e^{-4} \right) = 1 - 13 e^{-4} = 0.762
\end{align*}
\]
---
**6-21. Radius Calculation for Probability Sphere for Hydrogen 1s Electron**
Calculate the radius of the sphere that encloses a 50% probability of finding a hydrogen 1s electron. Repeat the calculation for a 90% probability.
---
The probability that a 1s electron will be found within a distance \(D a_0\) of the nucleus is given by
\[
\begin{align*}
\text{prob}(D) &= \int_0^{2\pi} d\phi \int_0^{\pi} \sin \theta \, d\theta \int_0^{Da_0} dr](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2f6cfa1-15bf-4f18-84f8-c308fdc99132%2F70667cfe-acc0-4b1c-913f-f6628abd0a19%2Flrmebwj.png&w=3840&q=75)
Transcribed Image Text:**6-20. Probability Calculation for Hydrogen 1s Electron**
Calculate the probability that a hydrogen 1s electron will be found within a distance \(2a_0\) from the nucleus.
---
This problem is similar to Example 6-10. The wave function for the 1s orbital of hydrogen is:
\[
\psi_{100} = \frac{1}{\sqrt{\pi}} \left( \frac{1}{a_0} \right)^{3/2} e^{-\sigma}
\]
where \(\sigma = r/a_0\), and the probability that the electron will be found within a distance \(2a_0\) from the nucleus is
\[
\begin{align*}
\text{prob} &= \int_0^{2\pi} d\phi \int_0^{\pi} \sin \theta \, d\theta \int_0^{2a_0} dr \, r \left( \frac{1}{\pi} \left( \frac{1}{a_0} \right)^3 \right) e^{-2\sigma} \\
&= 4 \left( \frac{1}{a_0} \right)^{3} \int_0^{2a_0} dr \, r^2 e^{-2\sigma} = 4 \int_0^{2} d\sigma \, \sigma^2 e^{-2\sigma} \\
&= 4 \left( \frac{1}{4} - \frac{13}{4} e^{-4} \right) = 1 - 13 e^{-4} = 0.762
\end{align*}
\]
---
**6-21. Radius Calculation for Probability Sphere for Hydrogen 1s Electron**
Calculate the radius of the sphere that encloses a 50% probability of finding a hydrogen 1s electron. Repeat the calculation for a 90% probability.
---
The probability that a 1s electron will be found within a distance \(D a_0\) of the nucleus is given by
\[
\begin{align*}
\text{prob}(D) &= \int_0^{2\pi} d\phi \int_0^{\pi} \sin \theta \, d\theta \int_0^{Da_0} dr
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