Chapter 28, Problem 55GP

To determine

The lifetime of the excited state that produced this line.

Answer to Problem 55GP

  $\text{6}\text{.3}\text{×}{\text{10}}^{\text{-14}}\text{s}$

Explanation of Solution

Given info:

Wavelengths, ${\text{λ}}_{\text{1}}\text{=}\text{487}\text{nm}\text{=}\text{487}{\text{×10}}^{\text{-9}}\text{m}$ and ${\text{λ}}_{\text{2}}\text{=}\text{489}\text{nm}\text{=}\text{489}{\text{×10}}^{\text{-9}}\text{m}$

Formula used:

The energy of a photon is $\text{E = }\frac{\text{hc}}{\mathrm{\lambda }}$

  $\text{λ}$ is wavelength.

  $\mathrm{h}$ is Planck’s constant, $6.63\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}$

  $\mathrm{c}$ is speed of light, $3.0\times {10}^8\mathrm{m}/\mathrm{s}$

  $\text{E}$ is energy.

Calculation:

First calculate the difference in measured energies:

  $\text{ΔE}\text{=}{\text{E}}_{\text{2}}{\text{- E}}_{\text{1}}$

  $\text{=}\frac{\text{hc}}{{\text{λ}}_{\text{2}}}\text{-}\frac{\text{hc}}{{\text{λ}}_{\text{1}}}\mathrm{...}\left(1\right)$

Substitute all the values in equation $\left(1\right)$

  $\text{=}\left(\text{6}\text{.63}\times {\text{10}}^{\text{-34}}\text{J}\cdot \text{s}\right)\left(\text{3}\text{.00}{\text{×10}}^{\text{8}}\text{m/s}\right)\left(\frac{\text{1}}{\text{489}{\text{×10}}^{\text{-9}}\text{m}}\text{-}\frac{\text{1}}{\text{487}{\text{×10}}^{\text{-9}}\text{m}}\right)$

  $\text{=1}\text{.67}\text{×}{\text{10}}^{\text{-21}}\text{J}$ (neglect the negative sign)

Now calculate the lifetime of the excited state by $\text{ΔEΔt}\approx \mathrm{\hslash }$

  $\left(1.67\times {10}^{-21}\text{J}\right)\mathrm{\Delta }\text{t }\approx \text{1}\text{.055}\times {\text{10}}^{\text{-34}}\text{J}\cdot \text{s}$

  $\text{Δt}=6.3\times {\text{10}}^{\text{-14}}\text{s}$

Conclusion:

The lifetime of the excited state that produced this line is $6.3\times {\text{10}}^{\text{-14}}\text{s}$ .