Question

In $L{i}^{++}$ electron in first Bohr orbit is excited to a level by a radiation of wavelength $X$. 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 $\lambda$ ?

• A. $11.4\text{nm}$
• B. $12.3\text{nm}$
• C. $10.8\text{nm}$
• D. $9.4\text{nm}$

Answer 1

The correct option is C $10.8\text{nm}$

Spectral lines obtained on account of trnsition from $n^{th}$ orbit to various lower orbits is $=\frac{n(n-1)}{2}$

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

$\Rightarrow n=4$

$\Delta E=\frac{hc}{\lambda }=\frac{-Z^2}{n^2}\left(13.6eV\right)$

$\Rightarrow \frac{1}{\lambda }=Z^2\left(\frac{13.6eV}{hc}\right)(\frac{1}{n_2^2}-\frac{1}{n_1^2})$

$\Rightarrow \frac{1}{\lambda }=\frac{\left(13.6\right)(3{)}^2}{hc}[1-\frac{1}{16}]eV$

$\Rightarrow \lambda =\frac{1242\times 16}{\left(13.6\right)\times \left(9\right)\left(15\right)}\text{nm}\approx 10.8\text{nm}$.

Hence option $\left(D\right)$ is correct.