Question

In $ \, Li^{+ +}$ ,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} \, J \, s \, ;c=3\times 10^{8} \, m \, s^{- 1}$ )

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

Answer 1

Answer: (A)

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