An electron in the hydrogen atom has a wave function given by $\psi(r,\theta,\phi) = \left(\frac{1}{a_0\sqrt{\pi}}\right)^{3/2} \exp[-r^2/(2a_0^2)]$. What is the probability that a measurement of the electron's energy yields the ground state energy of hydrogen? (Note: this will require an integral that needs to be done numerically.)
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The ground state energy of hydrogen is given by the formula: E = -13.6 eV / n^2 where n is the principal quantum number. For the ground state, n = 1, so the ground state energy is: E = -13.6 eV Show more…
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An electron in a hydrogen atom is in the ground state (1s). 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: $\psi_{1 s}(r)=A_{1 s} e^{-r / a_{0}}=e^{-r / a_{0}} / \sqrt{\pi a_{0}^{3}}$.
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