Answer
See answers.
Work Step by Step
Use equation 27-4 to calculate the difference in measured energies for the two extreme wavelengths.
$$\Delta E = E_2-E_1=\frac{hc}{\lambda_2}-\frac{hc}{\lambda_1}$$
$$\Delta E = (1240 eV \cdot nm)(\frac{1}{487 nm}-\frac{1}{489nm})(1.60\times10^{-19}J/eV)$$
$$\Delta E = 1.67\times10^{-21}J$$
Estimate the lifetime of the excited state using equation 28–2. Assume that the energy width is the uncertainty.
$$\Delta E \Delta t \geq \hbar$$
$$\Delta t \geq \frac{\hbar}{\Delta E}=\frac{1.055\times10^{-34}J \cdot s}{1.67\times10^{-21}J }\approx 6.32 \times10^{-14}s$$
