Question

In $L{i}^{++}$ ,electron in first Bohr orbit is excited to a level by a radiation of wavelength $\lambda$ .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 $\lambda$ ?
(Given: $h=6.63\times 10^{-34}Js;c=3\times 10^{8}m{s}^{-1}$ )

• A. $10.8nm$
• B. $9.4nm$
• C. $11.4nm$
• D. $12.3nm$

Answer 1

Answer: (A)

Total Number of possible spectral lines that can be emmited when atom de-excites from state $n$ to groundstate is $\_{}^n{C}_2^{}$
$\therefore {\left(\right)}^n{C}_2=6\Rightarrow \frac{n(n-1)}{2}=6\Rightarrow n=4$
$\frac{1}{\lambda }=R{z}^2\left(\frac{1}{1^2}-\frac{1}{4^2}\right)=1.097\times {\left(10\right)}^7\times 9\times \left(1-\frac{1}{16}\right)$
$\Rightarrow \lambda =10.8nm$