The difference between the radii of n^{text{t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The radius of the $n^{	ext{th}}$ orbit of a hydrogen atom is given by $r_n = r_0 n^2$,where $r_0$ is the Bohr radius.Given the condition: $r_{n+1

Vedclass · Accepted Answer

$\frac{2h}{\pi}$

Vedclass · Answer

(B) The radius of the $n^{	ext{th}}$ orbit of a hydrogen atom is given by $r_n = r_0 n^2$,where $r_0$ is the Bohr radius.Given the condition: $r_{n+1} - r_n = r_{n-1}$.Substituting the formula: $r_0(n+1)^2 - r_0 n^2 = r_0(n-1)^2$.Dividing by $r_0$: $(n+1)^2 - n^2 = (n-1)^2$.Expanding the terms: $(n^2 + 2n + 1) - n^2 = n^2 - 2n + 1$.Simplifying: $2n + 1 = n^2 - 2n + 1$.Rearranging: $n^2 - 4n = 0$,which gives $n(n-4) = 0$.Since $n$ must be a positive integer,$n = 4$.The angular momentum $L$ of an electron in the $n^{	ext{th}}$ orbit is given by Bohr's quantization condition: $L = \frac{nh}{2\pi}$.Substituting $n = 4$: $L = \frac{4h}{2\pi} = \frac{2h}{\pi}$.

The difference between the radii of $n^{\text{th}}$ and $(n+1)^{\text{th}}$ orbits of a hydrogen atom is equal to the radius of the $(n-1)^{\text{th}}$ orbit of hydrogen. The angular momentum of the electron in the $n^{\text{th}}$ orbit is $.........$ ($h$ is Planck's constant).

Similar Questions

The following parameter is the same for all hydrogen-like atoms and ions in their ground state.

The ground state energy of a hydrogen atom is $-13.6 \ eV$. In this state,the potential energy is . . . . . . $eV$.

Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?

When a hydrogen atom emits a photon of energy $12.09 \,eV$,its orbital angular momentum changes by (where $h$ is Planck's constant)

Vedclass Test Series

Exam Paper Generator

Online Exam Module