Question

In $Li^{++}$, electron in first 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 : $h = 6.63 \times 10^{-34} \; Js ; c = 3 \times 10^8 \; ms^{-1})$ ;

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

Answer 1

Answer: (C)

10.8 nm

Answer 2

$\Delta E =\frac{hc}{\lambda}$
$13.6\times9-0.85 \times9 = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{ 9 \times\left(13.6 -0.85\right)eV}$
$= \frac{1240eV .nm}{9\times12.75 eV}$
$\lambda = 10.8\, nm$