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A sample of hydrogen atom is excited to n=4 state. Ratio of number of lines in the ultraviolet and visible region are?
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A sample of hydrogen atom is excited to n=4 state. Ratio of number of ...
Ratio of number of lines in the ultraviolet and visible region for an excited hydrogen atom in the n=4 state:

When a hydrogen atom is excited, its electron moves to a higher energy level or shell. The energy levels in a hydrogen atom are denoted by the principal quantum number (n), where n=1 represents the ground state.

In this case, the hydrogen atom is excited to the n=4 state. The electron in this state has a higher energy compared to the ground state, and it can subsequently transition to lower energy levels by emitting photons.

Understanding Energy Levels and Transitions
The energy levels of a hydrogen atom can be calculated using the formula:

E = -13.6 eV / n²

Where E is the energy level and n is the principal quantum number. The energy difference between two energy levels can be calculated by taking the difference of their energies:

ΔE = E₂ - E₁

When an electron transitions between energy levels, it emits or absorbs a photon with energy equal to the energy difference between the levels. The energy of a photon can be calculated using the formula:

E = h * c / λ

Where E is the energy of the photon, h is the Planck's constant (6.626 x 10⁻³⁴ J s), c is the speed of light (3 x 10⁸ m/s), and λ is the wavelength of the photon.

Determining the Ratio of Lines in the Ultraviolet and Visible Region

To determine the ratio of lines in the ultraviolet and visible region for an excited hydrogen atom in the n=4 state, we need to consider the possible transitions and the corresponding wavelengths of the emitted photons.

Transitions from n=4 to lower energy levels:

1. Transition from n=4 to n=3:
- ΔE = E₃ - E₄ = (-13.6 eV / 3²) - (-13.6 eV / 4²) = 1.51 eV
- E = (1.51 eV) * (1.6 x 10⁻¹⁹ J/eV) = 2.42 x 10⁻¹⁹ J
- λ = (6.626 x 10⁻³⁴ J s * 3 x 10⁸ m/s) / (2.42 x 10⁻¹⁹ J) = 8.23 x 10⁻⁷ m (infrared region)

2. Transition from n=4 to n=2:
- ΔE = E₂ - E₄ = (-13.6 eV / 2²) - (-13.6 eV / 4²) = 10.2 eV
- E = (10.2 eV) * (1.6 x 10⁻¹⁹ J/eV) = 16.32 x 10⁻¹⁹ J
- λ = (6.626 x 10⁻³⁴ J s * 3 x 10⁸ m/s) / (16.32 x 10⁻¹⁹ J) = 3.82 x 10⁻⁷ m (visible region)

3. Transition from n=4 to n
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