Question

The Balmer series of hydrogenic spectral lines refers to an electron transitioning from \(n\geq 3\) to \(n=2\). A hydrogenic atom with atomic number \(Z=24\) undergoes a Balmer transition of the largest possible wavelength. The emitted photon has energy \(E_0\). Another hydrogenic atom with atomic number \(Z=25\) undergoes a similar Balmer transition with emitted photon energy \(E_1\). Then \(|E_1-E_0|\), in eV, is closest to:

• A. 9
• B. 27
• C. 54
• D. 90

Hint:

For hydrogenic atoms, transition energies vary as \(Z^2\). Even a small increase in atomic number can produce a large change in emitted photon energy.

Answer 1

The Correct Option is D

Step 1: Pin down which Balmer line is meant.
In the Balmer series the electron falls to $n_f = 2$. The largest possible wavelength means the smallest emitted energy, and the smallest energy jump is from the nearest higher level, $n_i = 3$. So both atoms make the $3 \rightarrow 2$ transition.
Step 2: Write the energy of that line using the Rydberg energy.
For a hydrogen-like atom the photon energy is \[ E = R_\infty Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] with $R_\infty = 13.6$ eV.
Step 3: Collect the constant factor once.
The bracket is the same for both atoms: \[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \] so define the common factor \[ k = 13.6 \times \frac{5}{36} = \frac{68}{36} \approx 1.889 \text{ eV} \] and then $E = k Z^2$.
Step 4: Avoid computing each energy separately.
Since $E_0 = k(24)^2$ and $E_1 = k(25)^2$, the difference factors neatly: \[ |E_1 - E_0| = k\left(25^2 - 24^2\right) \]
Step 5: Use the difference-of-squares shortcut.
\[ 25^2 - 24^2 = (25 - 24)(25 + 24) = 1 \times 49 = 49 \] so \[ |E_1 - E_0| = 1.889 \times 49 \]
Step 6: Evaluate and round to the nearest option.
\[ |E_1 - E_0| \approx 92.6 \text{ eV} \] which is closest to 90 eV. The factoring trick gives the same value as the key with far less arithmetic.
\[ \boxed{|E_1 - E_0| \approx 90 \text{ eV}} \]