Question

The first three energy levels of the fictitious element X were shown in FIGURE P38.56. An electron with a speed of 1.4 X 10⁶ m/s collides with an atom of element X. Shortly afterward, the atom emits a photon with a wavelength of 1240 nm. What was the electron's speed after the collision? Assume that, because the atom is much more massive than the electron, the recoil of the atom is negligible. Hint: The energy of the photon is not the energy transferred to the atom in the collision.

Short answer

The electron's speed after collision is $6.14\times {10}^5\frac{m}{s}$

Step-by-step solution

Given information

The first three energy levels of the fictitious element X were shown in FIGURE P38.56. An electron with a speed of 1.4 X 10⁶ m/s collides with an atom of element X. Shortly afterward, the atom emits a photon with a wavelength of 1240 nm.

Explanation

First we determine the energy of the photon:

$E_{ph}=\frac{hc}{\lambda }$

Substitute the values:

$$\begin{gathered} E_{ph}=\frac{6.63·{10}^{-34}·3·{10}^8}{1240·{10}^{-9}} \\ {E}_{ph}=1eV \end{gathered}$$

This energy corresponds 3$\to$2 transition, so after collision atom is in the n = 3 state. Before collision atom was in ground state (n = 1).|

Calculations:

Energy which electron loses is:

$\Delta E_{13}=E_3-E_1$

Substitute the values:

$$\begin{gathered} \Delta E_{13}=-2\left(eV\right)+6.5\left(eV\right) \\ \Delta E_{13}=4.5\mathrm{eV} \end{gathered}$$

$\Delta E_k=\Delta E_{13}=4.5\mathrm{eV}$

The lost kinetic energy is given as:

$-\Delta E_k=E_{kf}-E_{ki}$

Express in terms of E_kf

$E_{kf}=E_{ki}-\Delta E_k$

The expression for kinetic energy is:

$\frac{m{v}_f^2}{2}=\frac{m{v}_i^2}{2}-\Delta E_k$

Express in terms of v_f:

_{$v_f=\sqrt{\frac{2·\left(\frac{m{v}_i^2}{2}-\Delta E_k\right)}{m}}$}

Substitute the values:

$$\begin{gathered} v_f=\sqrt{\frac{2·\left(\frac{9.11·{10}^{-31}·{\left(1.4·{10}^6\right)}^2}{2}-4.5·1.6·{10}^{-19}\right)}{9.11·{10}^{-31}}} \\ {v}_f=6.14·{10}^5\frac{m}{s} \end{gathered}$$