Question

Hydrogen gas atoms absorb radiations of wavelength ${\lambda }_0$ and consequently emit radiations of 6 different wavelengths of it, two wavelengths are shorter than ${\lambda }_0$. Choose the correct alternatives.

• A. The final excited state of the atoms in n = 3.
• B. The final excited state of the atoms is n = 4
• C. The initial state of the atom is n = 2
• D. Initial state of the atoms is n = 3

Answer 1

Answer: (A), (C)

The correct options are
A The final excited state of the atoms in n = 3.
C The initial state of the atom is n = 2

$\begin{array}{c}\text{if electron excites to a Energy level, no of spectral}\end{array}$

$\text{lines}=\frac{n(n-1)}{2}$

$\begin{array}{c}n=\text{no of orbit to}\\ \text{which excite)}\end{array}$

$\begin{array}{c}\text{Given Spectral line}=6\\ \begin{array}{c}\frac{n(n-1)}{2}=6\\ n=4(3^{rd\text{excited state})}\end{array}\end{array}$

$\begin{array}{c}2\left(\lambda \right)\text{is shorter than}{x}_0\text{mean have less}\\ \text{energy}\begin{array}{cc}n=4& \to n=3\\ n=3& \to n=2\end{array}\end{array}$

$\text{So, initial state will be}2$

$\text{and,}\begin{array}{cc}n& =4\to n=1\\ n& =3⟶n=1\\ n& =2⟶n=1.\end{array}$

$\text{lie in lymen lemen}$

$\text{option a, and}c\text{are. correct.}$