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: Understanding the Question:
The question asks for the wavelength of the first spectral line of the Pfund series in the hydrogen emission spectrum.
We are given the Rydberg constant and need to apply the Rydberg formula.

Step 2: Key Formula or Approach:
The Rydberg formula for the wavelength of spectral lines in the hydrogen spectrum is:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where $R$ is the Rydberg constant ($1.097 \times 10^7$ m$^{-1}$).
For the Pfund series, the electron transitions end in the $n_1 = 5$ energy level.
The first spectral line corresponds to the transition from the adjacent higher energy level, which is $n_2 = 6$.

Step 3: Detailed Explanation:

• Substitute $n_1 = 5$ and $n_2 = 6$ into the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{5^2} - \frac{1}{6^2} \right) \] \[ \frac{1}{\lambda} = R \left( \frac{1}{25} - \frac{1}{36} \right) \] \[ \frac{1}{\lambda} = R \left( \frac{36 - 25}{25 \times 36} \right) \] \[ \frac{1}{\lambda} = R \left( \frac{11}{900} \right) \]

• Solving for the wavelength $\lambda$:
\[ \lambda = \frac{900}{11 R} \]

• Substituting the value of $R = 1.097 \times 10^7$ m$^{-1}$:
\[ \lambda = \frac{900}{11 \times 1.097 \times 10^7} \] \[ \lambda = \frac{900}{12.067 \times 10^7} \approx 74.58 \times 10^{-7}\text{ m} \]

• Convert the wavelength from meters to nanometers ($1\text{ nm} = 10^{-9}\text{ m}$):
\[ \lambda \approx 7458 \times 10^{-9}\text{ m} = 7458\text{ nm} \]

Step 4: Final Answer:
The wavelength of the first spectral line of the Pfund series is 7458 nm.