Hydrogen -like ion

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Question

$\begin{array}{l} A doubly ionized {Li}^{2 +} ion in the ground state absorbs 108.8 eV of energy. \\ Find the increase in angular momentum of electron. (take h = 6 .63 \times 10^{- 34} J - s) \end{array}$

Moderate

Solution

$\begin{array}{l} E_{n} = \frac{- 13.6}{n^{2}} Z^{2} \\ Z = 3, E_{n} = \frac{- 122.4}{n^{2}} \\ E_{1} = - 122.4 e V \\ E_{2} = - 30.6 e V \\ E_{3} = - 13.6 e V \\ E_{3} - E_{1} = - 13.6 - (- 122.4) = 108.8 e V \\ Change in angular momentum \\ = \frac{3 h}{2 π} - \frac{h}{2 π} \\ = 2 (\frac{h}{2 π}) \\ = 2.11 \times 10^{- 34} \end{array}$

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Similar Questions

In a sample of ions, a Li⁺⁺ ion electron in first Bohr orbit is excited to a level by a radiation of wavelength λ. When the ion gets de-excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of λ ?

$(Given, h = 6.63 \times 10^{- 34} Js, c = 3 \times 10^{8} {ms}^{- 1})$