Question

The energy of electron in the \(\mathrm{n}^{\text {th }}\) orbit of hydrogen atom is expressed as \(E_{n}=-\left[(13.6) / \mathrm{n}^{2}\right] e v .\) The shortest and longest wave length of lyman series will be. (A) \(912 \bar{\AA}, 1216 \AA\) (B) \(1315 \AA, 1530 \AA\) (C) \(5463 \AA, 7858 \AA\) (D) None of these

Short answer

The shortest and longest wavelengths of the Lyman series are approximately \(912 \mathring{A}\) and \(1216 \mathring{A}\), respectively, which corresponds to answer choice (A).

Step-by-step solution

Calculate the energy difference between the energy levels

For the shortest and longest wavelength in Lyman series, we consider the transitions from orbit n = 2 to n = 1 (longest wavelength) and n = ∞ to n = 1 (shortest wavelength). Let's find the change in energy for these transitions: For n = 2 to n = 1: ΔE₁ = |E₂ - E₁| = \(\left|\frac{-13.6}{2^2} - \frac{-13.6}{1^2}\right|\) eV For n = ∞ to n = 1: ΔE₂ = |E_∞ - E₁| = \(\left|\frac{-13.6}{\infty^2} - \frac{-13.6}{1^2}\right|\) eV

Convert energy difference to wavelength

Next, we need to convert the energy differences ΔE₁ and ΔE₂ to wavelengths using the formula: ΔE = h\(\frac{c}{\lambda}\), where ΔE is the change in energy, h is Planck's constant (4.136 × 10⁻¹⁵ eV·s), and c is the speed of light (3 × 10⁸ m/s). Longest wavelength (n = 2 to n = 1): \(\lambda_1 = \frac{hc}{\Delta E_1}\) Shortest wavelength (n = ∞ to n = 1): \(\lambda_2 = \frac{hc}{\Delta E_2}\)

Calculate the wavelengths and convert to Ångströms

Calculate the shortest and longest wavelengths using the energy differences obtained in step 1: Longest wavelength (n = 2 to n = 1): \(\lambda_1 = \frac{(4.136 \times 10^{-15})(3 \times 10^8)}{\Delta E_1}\) Shortest wavelength (n = ∞ to n = 1): \(\lambda_2 = \frac{(4.136 \times 10^{-15})(3 \times 10^8)}{\Delta E_2}\) Finally, convert these wavelengths from meters to Ångströms by multiplying by \(10^{10}\): \(\lambda_1 \times 10^{10} = \) Longest wavelength in Ångströms \(\lambda_2 \times 10^{10} = \) Shortest wavelength in Ångströms After calculating the values, we find that the shortest and longest wavelengths of the Lyman series are approximately \(1216 \mathring{A}\) and \(912 \mathring{A}\), respectively. This corresponds to the answer choice (A) \(912 \mathring{A}, 1216 \mathring{A}\).

Key concepts

hydrogen atom

The hydrogen atom is the simplest type of atom and consists of just one proton and one electron. Because it's the most basic atom, it's often used to introduce atomic structure concepts.
The electron in a hydrogen atom orbits around the proton in distinct paths called orbits or energy levels.

• These orbits are quantized, meaning the electron can only occupy certain allowed energy levels.
• When an electron jumps between these levels, it either absorbs or emits energy in the form of light.

Understanding these transitions is crucial to studying atomic spectra, like the Lyman series, where the electron moves between higher energy levels and the ground state.

energy levels

In the hydrogen atom, energy levels, also known as electron orbitals, describe the quantized paths that electrons take around the nucleus. Each "n" in the formula E_n = - (13.6/n^2) corresponds to a different principal quantum number, representing an energy level of the electron.
These levels are critical because:

• Lower energy levels are closer to the nucleus and are more stable.
• Higher energy levels require more energy and are less stable.
• When an electron transitions from a higher to a lower energy level, it emits energy as electromagnetic radiation, resulting in observable spectra like the Lyman series.

The energy associated with these levels decreases as increases, creating a specific energy level structure that impacts the behavior and light emission of the atom.

wavelength calculation

To calculate the wavelength of the light emitted or absorbed during an electron transition, we use the formula ∆E = h(c/λ). This relates the change in energy ( ∆E) to the wavelength ( λ) of the emitted light. The steps involve:

• Calculating the energy difference between the initial and final energy levels.
• Using Planck's constant ( h) and the speed of light ( c) to find the wavelength.

The equation shows that the energy difference ∆E is inversely proportional to λ. This means larger energy differences result in shorter wavelengths, typical of higher frequency ultraviolet light in the Lyman series. To convert meters to Ångströms, a unit dynamic in spectroscopy, use:
λ( Å) = λ( m) × 10^10.

Planck's constant

Planck's constant is a vital component in quantum physics, acting as a link between the energy of photons and the frequency of electromagnetic waves. Denoted by (h), it has a value of approximately 4.136 × 10⁻¹⁵ eV·s in electron volts (eV), a unit convenient for atomic level calculations.
Here's why it's key:

• In the equation ∆E = h(c/λ), it determines how energy is quantized in terms of frequency and wavelength.
• It allows us to calculate transitions like those seen in the hydrogen atom's Lyman series, where the emitted light corresponds to specific wavelength bands.

Planck's constant underscores the fundamental concept of quantized energy levels, forming a bridge from classic to quantum physics and allowing a deeper understanding of atomic emissions.