Question

Find the longest wavelength of the Paschen series for hydrogen atom. (Rydberg constant = 10⁷ /m)

• A. 2.06μm
• B. 20.6μm
• C. 4.86μm
• D. 48.6μm

Answer 1

The Correct Option is A

To find the longest wavelength of the Paschen series for the hydrogen atom, we use the Rydberg formula for hydrogen: 

\[\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\]

where:

• \(R = 10^7 \, \text{m}^{-1}\) is the Rydberg constant.
• \(n_1 = 3\) for the Paschen series.
• \(n_2\) is the principal quantum number for the higher energy level (where \(n_2 > n_1\)).

The longest wavelength corresponds to the smallest energy difference, which occurs when \(n_2 = n_1 + 1 = 4\).

Substituting these values into the Rydberg formula:

\[\frac{1}{\lambda} = 10^7 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)\]

Calculating the values inside the parentheses:

\[\frac{1}{3^2} = \frac{1}{9}\]

 and 

\[\frac{1}{4^2} = \frac{1}{16}\]

Therefore, we have:

\[\frac{1}{\lambda} = 10^7 \left( \frac{1}{9} - \frac{1}{16} \right) = 10^7 \left( \frac{16 - 9}{144} \right)\]\[\frac{1}{\lambda} = 10^7 \times \frac{7}{144}\]

Thus,

\[\lambda = \frac{144}{7 \times 10^7}\]

Calculating the wavelength:

\[\lambda = 2.06 \times 10^{-6} \, \text{m} = 2.06 \, \mu \text{m}\]

Therefore, the correct answer is 2.06μm.