# Tag Info

1

An excited state electron may transition to any lower level. From n=4, the electon could go to n=3, n=2 or n=1. Of these 3 transitions, only n=4 to n=2 (wavelength 486nm) is visible light. n=4 to n=1 is ultraviolet and n=4 to n=3 is infrared. The wavelengths of the transitions are given by the Rydberg formula. ...

1

$\langle r\rangle_{n,l,m}=\frac{a_0n^2}{Z}[1+\frac{1}{2}(1-\frac{l(l+1)}{n^2})]$ Source:McQuarrie, Quantum Chemistry.

1

Recall that the expectation value for any quantity $Q = Q(\mathbf x)$ in the (normalized) state $\psi$ is $$\langle Q\rangle_\psi = \int_{\mathbb R^3}d^3\mathbf x\, |\psi(\mathbf x)|^2Q(\mathbf x).$$ Let $(r,\theta, \phi)$ be the usual spherical coordinates. If we choose $Q(\mathbf x) = r$ and an energy eigenstate \$\psi_{n, \ell, m}(\mathbf x) = ...

0

Urgje gave you the answer. In its basic form (Schrödinger), the Hamiltonian is time-independent, therefore the general theory will tell you how to write the general solution of the Schrödinger equation as the sum/integral of the solutions of the spectral equation weighed by time-dependent exponentials.

1

The hydrogen nucleus has exactly zero nuclear binding energy, for the reason you gave in your question. The nuclear binding energy is the energy it takes to separate all the nucleons in a nucleus from each other. Since there is only the one nucleon, it's already separated from any other nucleons. For the same reason, a bare neutron has zero nuclear binding ...

Top 50 recent answers are included