if x is the wave number of series limit of lymn series in He+ ion then the wave number of second line of lymn series in hydrogen

atom

To find the wave number of the second line of the Lyman series in the hydrogen atom, we first need to understand the concept of wave numbers and how they relate to the transitions in hydrogen and hydrogen-like ions like He⁺.

Understanding Wave Numbers

The wave number (denoted as \( \bar{\nu} \)) is defined as the number of wavelengths per unit distance, typically expressed in reciprocal centimeters (cm^-1). It is related to the energy of the photon emitted or absorbed during an electronic transition in an atom.

Lyman Series Overview

The Lyman series corresponds to electronic transitions in hydrogen where an electron falls from a higher energy level (n ≥ 2) to the ground state (n = 1). The wave number for these transitions can be calculated using the Rydberg formula:

\( \bar{\nu} = R_H \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \)

Here, \( R_H \) is the Rydberg constant for hydrogen, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \) or \( 1.097 \times 10^5 \, \text{cm}^{-1} \).

Calculating the Wave Number for the Second Line

In the Lyman series, the second line corresponds to the transition from n = 3 to n = 1. Plugging these values into the Rydberg formula gives:

• For n = 3: \( \frac{1}{n^2} = \frac{1}{3^2} = \frac{1}{9} \)
• For n = 1: \( \frac{1}{1^2} = 1 \)

Substituting these into the formula:

\( \bar{\nu} = R_H \left( 1 - \frac{1}{9} \right) = R_H \left( \frac{8}{9} \right) \)

Calculating this gives:

\( \bar{\nu} = 1.097 \times 10^5 \, \text{cm}^{-1} \times \frac{8}{9} \approx 9.73 \times 10^4 \, \text{cm}^{-1} \)

Comparing with He⁺ Ion

For the He⁺ ion, which is a hydrogen-like atom, the Rydberg formula is modified to account for the nuclear charge (Z). The formula becomes:

\( \bar{\nu} = R_H Z^2 \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \)

For He⁺, Z = 2. If we want to find the wave number of the series limit (which corresponds to n approaching infinity), we can use n = ∞:

\( \bar{\nu} = R_H \cdot 2^2 \left( 1 - 0 \right) = 4 R_H \)

Calculating this gives:

\( \bar{\nu} = 4 \times 1.097 \times 10^5 \, \text{cm}^{-1} \approx 4.39 \times 10^5 \, \text{cm}^{-1} \)

Final Thoughts

In summary, the wave number of the second line of the Lyman series in hydrogen is approximately \( 9.73 \times 10^4 \, \text{cm}^{-1} \), while the wave number of the series limit for the He⁺ ion is about \( 4.39 \times 10^5 \, \text{cm}^{-1} \). This illustrates how the increased nuclear charge in He⁺ leads to higher energy transitions compared to hydrogen.