Question

In hydrogen type atom, shortest wavelength in Lyman series is given as 91 nm. Then the longest wavelength in Paschen series of this atom shall be

• A. 31.82 nm
• B. 113.3 nm
• C. 1.87 μm
• D. 2.31 μm

Hint:

The longest wavelength in a series corresponds to the transition between the lowest possible \( n_2 \) and \( n_1 \).

Answer 1

The Correct Option is C

To solve this problem, we need to understand the concept of spectral series in hydrogen and hydrogen-like atoms. Each series in the hydrogen spectrum corresponds to electron transitions from higher energy levels to a lower fixed energy level. Let us examine the two series mentioned: the Lyman series and the Paschen series.

1. Lyman Series: This series results from electronic transitions from higher energy levels to the first energy level (n=1). The shortest wavelength corresponds to the transition from infinity to level 1 (n = ∞ to n = 1). Given: Shortest wavelength in Lyman series is 91 nm.
2. Paschen Series: This series results from electronic transitions from higher energy levels to the third energy level (n=3). The longest wavelength corresponds to the transition from level 4 to level 3 (n = 4 to n = 3).

We use Rydberg's formula to calculate the wavelength (\lambda) in each transition:

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

• \lambda is the wavelength,
• R is the Rydberg constant, approximately 1.097 \times 10^7 \text{ m}^{-1},
• n_1 and n_2 are the principal quantum numbers of the lower and higher energy levels, respectively.

Calculate the longest wavelength in the Paschen series:

For the longest wavelength in the Paschen series (transition from n_2 = 4 to n_1 = 3), substitute into the Rydberg formula:

\frac{1}{\lambda} = R \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R \left(\frac{1}{9} - \frac{1}{16}\right)

\frac{1}{\lambda} = R \left(\frac{16 - 9}{144}\right) = R \left(\frac{7}{144}\right)

Substitute the value of R:

\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{7}{144}

\lambda = \frac{144}{7 \times 1.097 \times 10^7} \approx 1.87 \times 10^{-6} \text{ m}

Thus, the longest wavelength in the Paschen series is approximately 1.87 μm, which matches the given correct option.