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The wavelength of Lyman series of hydrogen atom appears in:
A
visible region
B
far infrared region
C
ultraviolet region
D
infrared region
Detailed Solution
$\dfrac{1}{\lambda}=R\left(\dfrac{1}{1^2}-\dfrac{1}{n^2}\right)$ for $n=2,3,4,\ldots$. Calculating the range: $(\lambda_L)_{max}=\dfrac{4}{3R}=\dfrac{4}{3}\times912\ \text{Å}=1216$ Å, and $(\lambda_L)_{min}=\dfrac{1}{R}\approx912$ Å. The range of $\lambda$ is 912 Å to 1216 Å, which lies in the UV region.
