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Answer 1

Question:-

Name the series of lines observed in the emission spectrum of Hydrogen in which the minimum value of $\sf n_2 = 2$. Calculate the wavelength of first line of this series [$\sf R_H=109677{cm}^{-1}$ ]

Answer:-

The minimum value of n2 is equal to 2 only in the first Lyman series.

We know that,

$\sf \frac{1}{ \lambda} = r | \frac{1}{ {( n_{1}) }^{2} } - \frac{1}{ { (n_{2}) }^{2} } |$

Where,

$\sf{\lambda}$ is the wavelength, n1 is the ground state and n2 is the excited state and R_H is the Rydberg's Constant.

For 1st Line of Lyman series, the values of n1 and n2 are 1 , 2.

Hence,

$\sf \implies \: \frac{1}{ \lambda} = 109677 | \frac{1}{ {1}^{2} } - \frac{1}{ {2}^{2} } | \\ \\ \sf \implies \: \frac{1}{ \lambda} = 109677 | \frac{4 - 1}{4} | \\ \\ \implies\sf \: \frac{1}{ \lambda} = 91.8 \: {nm} \times \frac{3}{4} \\ \\ \sf \implies \: \lambda = \frac{91.8 \times 4}{3} \\ \\ \sf \implies \large \: \lambda \: = 122.4 \: nm(Approx.)$