Answer to Question #108332 in Quantum Mechanics for Shivi

Question

The transition rate of electrons from the first excited state of the hydrogen atom to
the ground state is ~108
/s. What is the minimum range of energies of the resulting
photons that are emitted?

Answer 1

The energy-time uncertainty principle says

$$\Delta E\Delta t\ge\hbar/2.$$

Hence, the minimum range of energies of the resulting photons that are emitted

$$\Delta E=\frac{\hbar}{2\Delta t}=\frac{1.05\times 10^{-34}}{2\times 10^{-8}}=5.25\times 10^{-27} \:\rm J.$$