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

Concept: For a hydrogen-like atom, \[ E_n=-\frac{13.6Z^2}{n^2}\,\text{eV} \] For Balmer series, the electron finally reaches \(n=2\). The largest wavelength corresponds to the smallest energy transition. Therefore, \[ n_i=3 \rightarrow n_f=2 \]

Step 1: Determine the transition energy formula. Photon energy emitted: \[ E=13.6Z^2 \left( \frac{1}{2^2}-\frac{1}{3^2} \right) \] \[ E=13.6Z^2 \left( \frac{5}{36} \right) \] \[ E=\frac{68}{36}Z^2 \] \[ E=1.8889\,Z^2 \]

Step 2: Calculate \(E_0\) for \(Z=24\). \[ E_0=1.8889(24)^2 \] \[ E_0=1.8889(576) \] \[ E_0\approx1088\,\text{eV} \]

Step 3: Calculate \(E_1\) for \(Z=25\). \[ E_1=1.8889(25)^2 \] \[ E_1=1.8889(625) \] \[ E_1\approx1180.6\,\text{eV} \]

Step 4: Compute the difference. \[ |E_1-E_0| = 1.8889(625-576) \] \[ = 1.8889(49) \] \[ \approx92.6\,\text{eV} \] Thus, \[ |E_1-E_0|\approx 93\,\text{eV} \] Correct option: \[ \boxed{90} \]