\(E_1 = E_0 \left(\frac{1}{1^2} - \frac{1}{2^2}\right)\)
\(E_1 = E_0 \times \frac{3}{4}\)
\(E_2 = E_0\)
\(∴\) \(\frac{E_1}{E_2} = \frac{3}{4}\)
So, the correct option is (A): 3:4
Given below are two statements:
Statement (I): A spectral line will be observed for a \(2p_x \rightarrow 2p_y\) transition.
Statement (II): \(2p_x\) and \(2p_y\) are degenerate orbitals.
In the light of the above statements, choose the correct answer from the options given below:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an electron making a transition from a high energy state to a lower energy state. The photon energy of the emitted photon is equal to the energy difference between the two states.
Read More: Atomic Spectra
The Rydberg formula is the mathematical formula to compute the wavelength of light.
\[\frac{1}{\lambda} = RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\]Where,
R is the Rydberg constant (1.09737*107 m-1)
Z is the atomic number
n is the upper energy level
n’ is the lower energy level
λ is the wavelength of light
Spectral series of single-electron atoms like hydrogen have Z = 1.