$R \propto n ^{\frac{1 }{ 3}}$ and $v \propto n ^{\frac{2 }{ 3}}$
$R \propto n ^{\frac{2 }{ 3}}$ and $v \propto n ^{\frac{1 }{ 3}}$
$E=\frac{3}{2}\left(\frac{n^{2} h^{2} F^{2}}{4 \pi^{2} m}\right)^{\frac{1 }{ 3}}$
The Correct Option is (B) and(C):
$R \propto n ^{\frac{2 }{ 3}}$ and $v \propto n ^{\frac{1 }{ 3}}$
\(E=\frac{3}{2}\left(\frac{n^{2} h^{2} F^{2}}{4 \pi^{2} m}\right)^{\frac{1 }{ 3}}\)
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:
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
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.