Atomic StructureEasy
Question
The wavelength of radiation emitted out in the transition n = 4 to n = 1 in Li2+ ion is
Options
A.$\frac{135R}{16}$
B.$\frac{16}{135R}$
C.$\frac{16R}{135}$
D.$\frac{135}{16R}$
Solution
$\frac{1}{\lambda} = Rz^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right) = R \times 3^{2}\left( \frac{1}{1^{2}} - \frac{1}{4^{2}} \right) = \frac{135R}{16}$
$\therefore\lambda = \frac{16}{135R}$
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