JEE MAIN - Chemistry (2022 - 25th July Evening Shift - No. 15)
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
10
Explanation
Maximum number of emission lines
$$ =\frac{\left(n_{2}-n_{1}\right)\left(n_{2}-n_{1}+1\right)}{2} $$
$$ \begin{aligned} &\mathrm{n}_{2}=5 \\ &\mathrm{n}_{1}=1 \\ &\Rightarrow \frac{(5-1)(5-1+1)}{2}=10 \end{aligned} $$
Hence maximum number of emission lines observed are $10 .$
$$ =\frac{\left(n_{2}-n_{1}\right)\left(n_{2}-n_{1}+1\right)}{2} $$
$$ \begin{aligned} &\mathrm{n}_{2}=5 \\ &\mathrm{n}_{1}=1 \\ &\Rightarrow \frac{(5-1)(5-1+1)}{2}=10 \end{aligned} $$
Hence maximum number of emission lines observed are $10 .$
Comments (0)
