
Hydrogen atom excited energy level from fundamental state to $n = 3$. The number of spectral lines, according to Bohr, is:
A. $4$
B. $3$
C. $1$
D.$2$
Answer
220.5k+ views
Hint: In order to answer the issue, keep in mind that the hydrogen spectrum operates according to certain rules, and that the quantity of spectral lines in a certain series reflects all conceivable transitions from an energy level higher than that level to the level represented by that series.
Formula used:
The following formula can be used to determine how many spectral lines or emission lines the electrons will produce when they fall from orbit $n$ to the ground state:
$\text{Number of spectral lines} = \dfrac{{n(n - 1)}}{2}$.
Here, $n$ is an orbit where the ground state electron begins to fall.
Complete step by step solution:
In the question, we have given that a hydrogen atom electron moves from the $n = 3$ state to the ground state. To calculate the number of maximum spectral lines or emission lines that will be produced when the electron transitions from the $n = 3$ to the ground state.
Using the formula for finding the total number of possible spectral lines,
$\text{Number of spectral lines} = \dfrac{{n(n - 1)}}{2}$
Now, substitute the value of $n$in the above formula, then we obtain:
$\text{Number of spectral lines} = \dfrac{{3(3 - 1)}}{2} \\
\Rightarrow \text{Number of spectral lines} = \dfrac{{3(2)}}{2} \\
\therefore \text{Number of spectral lines} = 3 \\$
Therefore, the excited electron of a hydrogen atom at $n = 3$ will form a maximum of $3$ emission lines before it drops to the ground state.
Hence, the correct option is B.
Note: It should be noted that the emission lines or spectral lines form when an electron moves from a higher orbit to a lower orbit, which indicates that the electron is losing or emitting energy. These lines, often referred to as emission lines or spectral lines, are produced when an electron emits energy. An electron moves to a higher energy level and absorbs some of it then resulting in an absorption spectrum or spectral absorption lines.
Formula used:
The following formula can be used to determine how many spectral lines or emission lines the electrons will produce when they fall from orbit $n$ to the ground state:
$\text{Number of spectral lines} = \dfrac{{n(n - 1)}}{2}$.
Here, $n$ is an orbit where the ground state electron begins to fall.
Complete step by step solution:
In the question, we have given that a hydrogen atom electron moves from the $n = 3$ state to the ground state. To calculate the number of maximum spectral lines or emission lines that will be produced when the electron transitions from the $n = 3$ to the ground state.
Using the formula for finding the total number of possible spectral lines,
$\text{Number of spectral lines} = \dfrac{{n(n - 1)}}{2}$
Now, substitute the value of $n$in the above formula, then we obtain:
$\text{Number of spectral lines} = \dfrac{{3(3 - 1)}}{2} \\
\Rightarrow \text{Number of spectral lines} = \dfrac{{3(2)}}{2} \\
\therefore \text{Number of spectral lines} = 3 \\$
Therefore, the excited electron of a hydrogen atom at $n = 3$ will form a maximum of $3$ emission lines before it drops to the ground state.
Hence, the correct option is B.
Note: It should be noted that the emission lines or spectral lines form when an electron moves from a higher orbit to a lower orbit, which indicates that the electron is losing or emitting energy. These lines, often referred to as emission lines or spectral lines, are produced when an electron emits energy. An electron moves to a higher energy level and absorbs some of it then resulting in an absorption spectrum or spectral absorption lines.
Recently Updated Pages
[Awaiting input: Please provide the content from "Ask AI Response," "Competitor 1," and "Competitor 2," so I can perform the analysis and synthesize the requested metadata and headings.]

Young’s Double Slit Experiment Derivation Explained

A square frame of side 10 cm and a long straight wire class 12 physics JEE_Main

The work done in slowly moving an electron of charge class 12 physics JEE_Main

Two identical charged spheres suspended from a common class 12 physics JEE_Main

According to Bohrs theory the timeaveraged magnetic class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Electromagnetic Waves and Their Importance

