MODERN BYU 8

Problem 8: Hydrogen-like Ions Analysis

1. Identify Final State from Spectral Lines:

The problem states that after absorbing the photons, the emission spectrum shows 10 distinct lines.

The number of emission lines from a state $n_{final}$ to the ground state is given by the formula:

$$ \text{Number of Lines} = \frac{n_{final}(n_{final}-1)}{2} $$ $$ 10 = \frac{n_{final}(n_{final}-1)}{2} \implies n_{final}(n_{final}-1) = 20 $$

Solving for $n_{final}$, we get $n_{final} = 5$.

2. Determine Initial State and Element (Z):

The ions were initially in an excited state $n_{initial}$ and absorbed a photon to reach $n_{final}=5$. The energy gap must match the incident photon energy.

• Incident Photon Energy: $E_{ph} = \frac{144 E_0}{225} = \left(\frac{12}{15}\right)^2 E_0 = \frac{16}{25} E_0$
• Absorption Formula: $\Delta E = Z^2 E_0 \left( \frac{1}{n_{initial}^2} – \frac{1}{n_{final}^2} \right)$

Substituting values:

$$ Z^2 E_0 \left( \frac{1}{n_{initial}^2} – \frac{1}{5^2} \right) = \frac{16}{25} E_0 $$ $$ Z^2 \left( \frac{1}{n_{initial}^2} – \frac{1}{25} \right) = \frac{16}{25} $$

We solve for integer values of $Z$ (atomic number) and $n_{initial}$ by trial and error:

Try $n_{initial} = 3$:

$$ Z^2 \left( \frac{1}{9} – \frac{1}{25} \right) = \frac{16}{25} $$ $$ Z^2 \left( \frac{25 – 9}{225} \right) = \frac{16}{25} $$ $$ Z^2 \left( \frac{16}{225} \right) = \frac{16}{25} $$ $$ \frac{Z^2}{9} = 1 \implies Z^2 = 9 \implies Z = 3 $$

This yields a valid integer solution. Thus:

• Element: $Z=3$ (Lithium)
• Initial State: $n=3$ (Second excited state)
• Final State: $n=5$ (Fourth excited state)