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

Question

Quizplus · Accepted Answer

The average time an atom spends in the excited state before emitting a photon and transitioning to a lower energy state can be estimated using the uncertainty principle, which relates the uncertainty in energy ($\Delta E$) to the uncertainty in time ($\Delta t$) through the equation $\Delta E \Delta t \approx \hbar$, where $\hbar$ is the reduced Planck's constant ($1.055 \times 10^{-34} \, \text{J} \cdot \text{s}$). The natural line width ($\Delta \lambda$) can be related to the uncertainty in energy using the relation $\Delta E = \frac{hc}{\lambda^2} \Delta \lambda$, 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 subsequently $\Delta t$. The calculation yields a time 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