A 440-nm spectral line is produced by a transition from an excited state to the ground state.The natural line width of the spectral line is 0.020 pm.The average time the atom spends in the excited state is closest to which of the following? ( h = 1.055 × 10^-34 J • s = 6.59 × 10^-16 eV • s) A) 2.5 × 10^-6 s B) 2.5 × 10^-7 s C) 2.5 × 10^-8 s D) 2.5 × 10^-9 s E) 2.5 × 10^-10 s

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Quizplus · Accepted Answer

The average time an atom spends in the excited state before emitting a photon can be estimated using the uncertainty principle, $\Delta E \Delta t \approx \hbar$, where $\Delta E$ is the energy width of the spectral line, $\Delta t$ is the lifetime of the excited state, and $\hbar$ is the reduced Planck's constant ($1.055 \times 10^{-34} \, \text{J} \cdot \text{s}$). The energy width $\Delta E$ can be related to the wavelength width $\Delta \lambda$ and the central wavelength $\lambda$ of the spectral line through the relation $\Delta E = hc \frac{\Delta \lambda}{\lambda^2}$, where $h$ is Planck's constant and $c$ is the speed of light. Given $\Delta \lambda = 0.020 \, \text{pm} = 0.020 \times 10^{-12} \, \text{m}$ and $\lambda = 440 \, \text{nm} = 440 \times 10^{-9} \, \text{m}$, we can calculate $\Delta E$ and then use the uncertainty principle to find $\Delta t$. The calculation yields a value on the order of $10^{-9} \, \text{s}$, making option D the closest match.

A 440-Nm Spectral Line Is Produced by a Transition from an Excited