A sodium atom emits a photon with wavelength $818nm$nm shortly after being struck by an electron. What minimum speed did the electron have before the collision?

The energy of emitted photon is,

$E=\frac{hc}{\lambda }$

Here, $h$is the Planck's constant, $c$is the speed of the light, and $\lambda$is the wavelength of the light.

Substitute $1240eV-nm$for $hc$and $499nm$for $\lambda$in the equation $E=\frac{hc}{\lambda }$and solve for $E$.

$E=\frac{1240eV·nm}{499nm}$

$=1.516eV$

From the energy spectrum of sodium, the energy of the $3d$state and $3p$state is as follows:

$$\begin{gathered} E_{3d}-E_{3p}=3.620eV-2.104eV \\ =1.516eV \end{gathered}$$

This energy difference is equal to the energy of emitted photon. So it is clear that the atom is excited from ground state to $3d$state. Hence, the minimum kinetic energy of the electron is equal to energy of the $3d$state.

$\frac{1}{2}m{v}^2=E_{3d}$

Here, $m$is the mass of an electron and $v$is the minimum speed of an electron. Rearrange the equation for $v$.

$v=\sqrt{\frac{2E_{3d}}{m}}$

Substitute $3.620eV$for $E_{3d}$and $9.11\times {10}^{-31}kg$for $m$in the above expression.

$$\begin{gathered} v=\sqrt{\frac{2\left(3.620eV\right)\left({\displaystyle \frac{1.60\times {10}^{-19}J}{1eV}}\right)}{9.11\times {10}^{-31}kg}} \\ =1.13\times {10}^{6}m/s \end{gathered}$$

Therefore, the minimum speed of an electron is$1.13\times {10}^{6}m/s$