An electron initially present in an excited s | vedclass

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

Question

(B) Let the electron be excited from state $n_1$ to $n_2$ by absorbing a photon of energy $E = E_{n_2} - E_{n_1}$.When returning to the ground state $

Vedclass · Accepted Answer

$2^{nd}$ excited state

Vedclass · Answer

(B) Let the electron be excited from state $n_1$ to $n_2$ by absorbing a photon of energy $E = E_{n_2} - E_{n_1}$.When returning to the ground state $(n=1)$ from $n_2$,the total number of photons emitted is given by $\frac{n_2(n_2-1)}{2} = 10$.Solving for $n_2$: $n_2^2 - n_2 - 20 = 0 \implies (n_2-5)(n_2+4) = 0$. Since $n_2 > 0$,we have $n_2 = 5$.The energy of the incident photon is $E_{incident} = E_5 - E_{n_1}$.There are $7$ transitions that have higher energy than the incident photon. These transitions must be from levels $n > n_1$ to levels $n \le n_1$.The number of photons emitted with energy less than or equal to the incident photon is $10 - 7 = 3$. These correspond to transitions from levels $n \le n_1$ to $n=1$.The number of such transitions is $\frac{n_1(n_1-1)}{2} = 3$.Solving for $n_1$: $n_1^2 - n_1 - 6 = 0 \implies (n_1-3)(n_1+2) = 0$. Thus,$n_1 = 3$.Since $n_1 = 3$,the electron was initially in the $2^{nd}$ excited state ($n=3$ is the $2^{nd}$ excited state).

An electron initially present in an excited state of $H$ atom is further excited to another energy level by an incident photon. It releases $10$ photons while coming back to the ground state,out of which $7$ have higher energy than the incident photon. The electron was initially present in:

Similar Questions

$A$ $12.75 \ eV$ electron beam is used to bombard gaseous hydrogen at room temperature. The wavelengths emitted belong to which of the following series?

Bohr's atomic model is applicable to which of the following?

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{h^{2}}{xma_{0}^{2}}$. The value of $10x$ is ........ . ($a_{0}$ is radius of Bohr's orbit) (Nearest integer) [Given : $\pi=3.14$]

The ratio of the area covered by the second orbital to the first orbital is

Vedclass Test Series

Exam Paper Generator

Online Exam Module