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Wave Properties of particles. Schrodinger Equation.

Problem 6.74

Making use of the uncertainty principle, evaluate the minimum permitted energy of an electron in a hydrogen atom and its corresponding apparent distance from the nucleus.

Reveal Answer
6.74. Taking into account that p∼Δp∼ℏ/Δrp \sim \Delta p \sim \hbar / \Delta r and Δr∼r,\Delta r \sim r, we get E=p2/2m−e2/r≈ℏ2/2mr2−e2/rE=p^{2} / 2 m-e^{2} / r \approx \hbar^{2} / 2 m r^{2}-e^{2} / r. From the condition dE/dr=0d E / d r=0 we find reff≈ℏ2/me2=53pm,Emin≈−me4/2ℏ2=r_{e f f} \approx \hbar^{2} / m e^{2}=53 \mathrm{pm}, E_{m i n} \approx-m e^{4} / 2 \hbar^{2}= =−13.6eV=-13.6 \mathrm{eV}
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