Answer to Question #108332 in Quantum Mechanics for Shivi

Question #108332
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?
1
Expert's answer
2020-04-08T10:32:28-0400

The energy-time uncertainty principle says

ΔEΔt/2.\Delta E\Delta t\ge\hbar/2.

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

ΔE=2Δt=1.05×10342×108=5.25×1027J.\Delta E=\frac{\hbar}{2\Delta t}=\frac{1.05\times 10^{-34}}{2\times 10^{-8}}=5.25\times 10^{-27} \:\rm J.

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