Answer
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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.
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