In Li++, electron in first {Bohr} orbit is excited to a level by

Question:

In $\mathrm{Li}^{++}$, electron in first $\mathrm{Bohr}$ orbit is excited to a level by a radiation of wavelength $\lambda$. When the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $\lambda$ ?

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

  1. (1) $11.4 \mathrm{~nm}$

  2. (2) $9.4 \mathrm{~nm}$

  3. (3) $12.3 \mathrm{~nm}$

  4. (4) $10.8 \mathrm{~nm}$


Correct Option: , 4

Solution:

(4) Spectral lines obtained on account of transition from

$\mathrm{n}^{\text {th }}$ orbit to various lower orbits is $\frac{n(n-1)}{2}$

$\Rightarrow 6=\frac{n(n-1)}{2}$

$\Rightarrow \mathrm{n}=4$

$\Delta E=\frac{h c}{\lambda}=\frac{-Z^{2}}{n^{2}}(13.6 \mathrm{eV})$

$\Rightarrow \frac{1}{\lambda}=Z^{2}\left(\frac{13.6 \mathrm{eV}}{h c}\right) \quad\left(\frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}}\right)$

$=(13.4)(3)^{2}\left[1-\frac{1}{16}\right] \mathrm{eV}$

$\Rightarrow \lambda=\frac{1242 \times 16}{(13.4) \times(9)(15)} \mathrm{nm} \simeq 10.8 \mathrm{~nm}$

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