Atomic StructureHard

Question

The wavelength of spectral line obtained in the spectrum of ${LI}^{2 +}$ ion, when the transition takes place between two levels whose sum is 4 and difference is 2 , is

Options

A.$2.28 \times 10^{- 7}\text{ }cm$
B.$2.28 \times 10^{- 6}\text{ }cm$
C.$1.14 \times 10^{- 7}\text{ }cm$
D.$1.14 \times 10^{- 6}\text{ }cm$

Solution

$n_{1} \rightarrow$ lower energy level

$n_{2} \rightarrow$ higher energy level

$${n_{1} + n_{2} = 4,\ n_{2} = 3 }{n_{2} - n_{1} = 2,n_{1} = 1 }$$Rydderg's formula :

$${\frac{1}{\lambda} = R_{H}Z^{2}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack }{\frac{1}{\lambda} = R_{H}(3)^{2}\left\lbrack \frac{1}{1^{2}} - \frac{1}{3^{2}} \right\rbrack }{\frac{1}{\lambda} = 8R_{H} }{\lambda = \frac{1}{8R_{H}} }{\lambda = \frac{1}{8 \times 1.1 \times 10^{5}} }{\lambda = \frac{1000}{8.8} \times 10^{- 8}\text{ }cm }{\lambda = 113.63 \times 10^{- 8}\text{ }cm }{\lambda \simeq 1.14 \times 10^{- 6}\text{ }cm}$$

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