The wave function for the (1 mathrm{~s}) state of an electron in the hydrogen atom is [psi_{1
Question:
The wave function for the \(1 \mathrm{~s}\) state of an electron in the hydrogen atom is
\[\psi_{1 \mathrm{~s}}(ho)=\frac{1}{\sqrt{\pi a_{0}^{3}}} e^{-ho / a_{0}}\]
where \(a_{0}\) is the Bohr radius. The probability of finding the electron in a region \(\mathcal{W}\) of \(\mathbf{R}^{3}\) is equal to
\[\iiint_{\mathcal{W}} p(x, y, z) d V\]
where, in spherical coordinates,
\[p(ho)=\left|\psi_{1 \mathrm{~s}}(ho)ight|^{2}\]
Use integration in spherical coordinates to show that the probability of finding the electron at a distance greater than the Bohr radius is equal to \(5 / e^{2} \approx 0.677\). (The Bohr radius is \(a_{0}=5.3 \times 10^{-11} \mathrm{~m}\), but this value is not needed.)
Fantastic news! We've Found the answer you've been seeking!
Step by Step Answer:
Related Book For
Question Posted: