When the excited electron of a H atom from n = 5 drops to the ground state, the maximum number of emission lines observed are _____________.
Answer (integer)
10
Solution
Maximum number of emission lines
<br/><br/>
$=\frac{\left(n_{2}-n_{1}\right)\left(n_{2}-n_{1}+1\right)}{2}$
<br/><br/>
$$
\begin{aligned}
&\mathrm{n}_{2}=5 \\
&\mathrm{n}_{1}=1 \\
&\Rightarrow \frac{(5-1)(5-1+1)}{2}=10
\end{aligned}
$$
<br/><br/>
Hence maximum number of emission lines observed are $10 .$
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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