Question

A certain atom has an energy state 3.5 eV above the ground state. When excited to this state, the atom remains for 2 ms, on average, before it emits a photon and returns to the ground state. (a) What are the energy and wavelength of the photon? (b) What is the smallest possible uncertainty in energy of the photon?

Short answer

$$\begin{gathered} \mathrm{a}){\mathrm{E}}_{\mathrm{\gamma }}=3.5\mathrm{eV}\mathrm{and}\mathrm{\lambda }=355\mathrm{nm} \\ \mathrm{b})∆\mathrm{E}=1.65\times {10}^{-10}\mathrm{eV} \end{gathered}$$

Step-by-step solution

Calculate energy and wavelength

The energy of the emitted photon equals the energy difference between the two levels; And since the excited state is above the ground state by $∆\mathrm{E}=3.5\mathrm{eV}$, so the energy of the emitted photon is ${\mathrm{E}}_{\mathrm{\gamma }}=3.5\mathrm{eV}$.

From equation 38.1, the energy of a photon of wavelength is given by:

$\mathit{E}\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\mathbf{\lambda }}$

Solving for $\lambda$and substituting for $E_{\gamma }$our value for , we get the wavelength of the emitted photon:

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{hc}}{{\mathrm{E}}_{\mathrm{\gamma }}} \\ \mathrm{\lambda }=\frac{4.136\times {10}^{-15}\times 3\times {10}^8}{3.5} \\ \mathrm{\lambda }=3.55\times {10}^{-7}\mathrm{m}=355\mathrm{nm} \end{gathered}$$

Calculate the uncertainty in energy

From equation 39.3, the uncertainty $\mathbf{∆}\mathit{E}$in the energy of a state that is occupied for a time is given by: $\mathbf{∆}\mathit{E}\mathbf{∆}\mathit{t}\mathbf{\ge }\frac{\sout{\mathbf{h}}}{\mathbf{2}}$

Substituting for$∆\mathrm{t}=2\times {10}^{-6}\mathrm{s}$ (the time during which the atom remains in its excited state), we get:

$$\begin{gathered} ∆\mathrm{E}\ge \frac{\sout{\mathrm{h}}}{2∆\mathrm{t}} \\ ∆\mathrm{E}\ge \frac{1.055\times {10}^{-34}}{2\times 2\times {10}^{-6}} \\ ∆\mathrm{E}\ge \frac{\sout{\mathrm{h}}}{2∆\mathrm{t}} \\ ∆\mathrm{E}\ge 2.64\times {10}^{-29}\mathrm{J} \end{gathered}$$

So, the smallest possible uncertainty in energy of the photon is:

$∆\mathrm{E}=2.64\times {10}^{-29}\mathrm{J}=1.65\times {10}^{-10}\mathrm{eV}$