Question

An electron accelerates through a $12.5V$ potential difference, starting from rest, and then collides with a hydrogen atom, exciting the atom to the highest energy level allowed. List all the possible quantum-jump transitions by which the excited atom could emit a photon and the wavelength (in nm) of each.

Short answer

The wavelength is${\lambda }_1=656.08nm$,${\lambda }_2=102.56nm$and${\lambda }_3=121.56nm$.

Step-by-step solution

Given Information

We need to find the emit a photon and the wavelength.

Simplify

The atom could emit a photon in the $3$possible quantum-jump transition:

role="math" localid="1650276055698" $3\to 2,3\to 1and2\to 1$

Now, for $3\to 2$:

$$\begin{gathered} E_1=-1.51eV-\left(-3.40eV\right) \\ {E}_1=-1.51eV+3.40eV \\ {E}_1=1.89eV \end{gathered}$$

for $3\to 1$:

$$\begin{gathered} E_2=-1.51eV-\left(-13.60eV\right) \\ {E}_2=-1.51eV+13.60eV \\ {E}_2=12.09eV \end{gathered}$$

for $2\to 1$:

$$\begin{gathered} E_3=-3.40eV-\left(-13.60eV\right) \\ {E}_3=-3.40eV+13.60eV \\ {E}_3=10.2eV \end{gathered}$$

Calculation

Now, finding the wavelength :

$$\begin{gathered} {\lambda }_1=\frac{hc}{E_1} \\ {\lambda }_1=\frac{12.40eV\times nm}{1.89eV} \\ {\lambda }_1=656.08nm \\ {\lambda }_2=\frac{hc}{E_2} \\ {\lambda }_2=\frac{12.40eV\times nm}{12.09eV} \\ {\lambda }_2=102.56nm \\ {\lambda }_3=\frac{hc}{E_3} \\ {\lambda }_3=\frac{12.40eV\times nm}{10.2eV} \\ {\lambda }_3=121.56nm \end{gathered}$$