An electron in an hydrogen like species is associated with wavelength of 6.67A wave number of lymans series limit of that species is 438708/cm the species is excited electrically to some higher quantum state which has 4.808eV energy more than second excited state

What is kinetic energy of electron

Which is that hydrogen like species

To tackle this question, we need to break it down into several parts: identifying the hydrogen-like species, calculating the kinetic energy of the electron, and understanding the relationship between energy levels and wavelengths in such systems. Let's dive into it step by step.

Identifying the Hydrogen-like Species

A hydrogen-like species is any atom or ion that has only one electron, similar to hydrogen. The wave number of the Lyman series limit provided is 438708 cm^-1. The Lyman series corresponds to transitions where electrons fall to the first energy level (n=1) from higher levels.

The wave number (ν) is related to the energy (E) of the photon emitted or absorbed during these transitions by the equation:

E = hν

where h is Planck's constant. The energy can also be expressed in terms of the Rydberg formula for hydrogen-like atoms:

ν = RZ²(1/n_f² - 1/n_i²)

Here, R is the Rydberg constant (approximately 109677 cm^-1), Z is the atomic number, n_f is the final energy level, and n_i is the initial energy level. For the Lyman series, n_f = 1.

Calculating the Atomic Number

Using the wave number given, we can rearrange the Rydberg formula to find Z:

• ν = RZ²(1/1² - 1/n_i²)
• Substituting ν = 438708 cm^-1 and R = 109677 cm^-1:

438708 = 109677Z²(1 - 1/n_i²)

To find Z, we can assume a transition from n_i = 2 (the second excited state) to n_f = 1:

1 - 1/2² = 3/4

Now substituting back:

438708 = 109677Z²(3/4)

Solving for Z gives:

Z² = (438708 * 4) / (109677 * 3) ≈ 5.04

Thus, Z ≈ 2.25, which suggests the species is Helium (He⁺), as it has an atomic number of 2.

Determining the Kinetic Energy of the Electron

Next, we need to find the kinetic energy of the electron. The problem states that the electron is excited to a higher quantum state with an energy of 4.808 eV more than the second excited state (n=2). The energy levels for a hydrogen-like atom can be calculated using the formula:

E_n = - (Z² * 13.6 eV) / n²

For n=2:

E₂ = - (2² * 13.6 eV) / 2² = -13.6 eV

Now, the energy of the excited state (let's call it n_f) is:

E_f = E₂ + 4.808 eV = -13.6 eV + 4.808 eV = -8.792 eV

To find the kinetic energy (KE) of the electron in the excited state, we can use the relationship:

KE = -E_f = 8.792 eV

Summary of Findings

In summary, the hydrogen-like species is He⁺, and the kinetic energy of the electron in the excited state is approximately 8.792 eV. This process illustrates the fascinating interplay between quantum mechanics and atomic structure, showcasing how energy levels and transitions define the behavior of electrons in atoms.