$R \propto n ^{\frac{1 }{ 3}}$ 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:
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ 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.