For all deexcitations that result in the visible spectrum of hydrogen, what is n ?
To determine the value of n for all deexcitations that result in the visible spectrum of hydrogen, we need to understand the energy levels and transitions in the hydrogen atom.
The visible spectrum of hydrogen corresponds to transitions in which electrons jump from higher energy levels to lower energy levels. These transitions involve the emission of photons with specific wavelengths that fall within the visible range.
The energy levels in hydrogen are described by the Rydberg formula:
1/λ = R * (1/n₁^2 - 1/n₂^2)
Where:
λ is the wavelength of the emitted photon,
R is the Rydberg constant (approximately 1.097 × 10^7 m⁻¹),
n₁ is the initial energy level (greater than n₂), and
n₂ is the final energy level (less than n₁).
Since we're interested in the visible spectrum, we know that the wavelength of the emitted photons will fall within a specific range, roughly 400 to 700 nanometers.
To find the value of n, we can start by examining the Balmer series, which contains the transitions corresponding to the visible spectrum. The Balmer series is defined by n₁ = 2, as the electrons transition from higher energy levels to the second energy level (n₂ = 2).
Plugging n₁ = 2 into the Rydberg formula, we can solve for n₂:
1/λ = R * (1/2^2 - 1/n₂^2)
We want to determine the possible values of n₂ that can give us wavelengths within the visible range. By substituting different values of n₂ and solving for λ, we can determine the corresponding visible wavelengths.
Note that for hydrogen, n₂ cannot be smaller than 1, as the energy level of the hydrogen atom is always a positive integer.
By systematically solving the equation for different values of n₂ greater than or equal to 1, we can identify the corresponding energy levels (n₁) and answer the question.