Question

In the pfund series of hydrogen spectrum, the wavelength of first spectral line is (Rydberg constant = 1.097 $\times 10^7$ m$^{-1}$)

• A. 8547 nm
• B. 6574 nm
• C. 3729 nm
• D. 7458 nm

Hint:

For any spectral series, the "first line" always has the minimum energy and maximum wavelength ($n_2 = n_1 + 1$).
The "series limit" or "last line" has the maximum energy and minimum wavelength ($n_2 = \infty$).

Answer 1

The Correct Option is D

Step 1: Convert the Rydberg constant into an energy scale instead of a wavenumber.
The quantity $hcR$ is the well known ground-state binding energy of hydrogen, $hcR \approx 13.6\text{ eV}$, so transition energies can be written directly in eV as $\Delta E = 13.6\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$.
Step 2: Identify the transition.
The Pfund series ends at $n_1 = 5$, and its first line comes from the very next level, $n_2 = 6$.
Step 3: Compute the transition energy. \[ \Delta E = 13.6\left(\frac{1}{25}-\frac{1}{36}\right) = 13.6\times\frac{11}{900} \approx 0.166\text{ eV} \]
Step 4: Convert this energy to wavelength using $hc/e = 1240$ eV-nm. \[ \lambda = \frac{1240}{0.166} \] \[ \boxed{\lambda \approx 7458\text{ nm}} \]