Question

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?

Short answer

Use the expression for the energy difference between two energy levels is related to the wavelength of photon.

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

Step-by-step solution

Energy of emitted photon:

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.

Substitution of values:

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:

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

Energy difference:

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$1.13\times {10}^{6}m/s$