Question
A hydrogen atom emits ultraviolet radiation of wavelength 102.5nm. What are the quantum numbers of the states involved in the transition?
✓
Answer
As the light emitted lies in ultraviolet range the line lies in hyman series.$\frac{1}{\lambda}=\text{R}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow\frac{1}{102.5\times10^{-9}}=1.1\times10^{7}\bigg(\frac{1}{1^2}-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow\frac{10^9}{102.5}=1.1\times10^7\bigg(1-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow\frac{10^2}{102.5}=1.1\times10^7\bigg(1-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow1-\frac{1}{\text{n}^2_2}=\frac{100}{102.5\times1.1}$
$\Rightarrow\frac{1}{\text{n}^2_2}=\frac{1-100}{102.5\times1.1}$
$\Rightarrow\text{n}_2=2.97=3$
$\Rightarrow\frac{1}{102.5\times10^{-9}}=1.1\times10^{7}\bigg(\frac{1}{1^2}-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow\frac{10^9}{102.5}=1.1\times10^7\bigg(1-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow\frac{10^2}{102.5}=1.1\times10^7\bigg(1-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow1-\frac{1}{\text{n}^2_2}=\frac{100}{102.5\times1.1}$
$\Rightarrow\frac{1}{\text{n}^2_2}=\frac{1-100}{102.5\times1.1}$
$\Rightarrow\text{n}_2=2.97=3$
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