Modern Physics Ques 7
- In $\mathrm{Li}^{++}$, 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 $\lambda$ ?
(Main 2019, 10 April II)
[Take, $h=6.63 \times 10^{-34} \mathrm{Js} ; c=3 \times 10^8 \mathrm{~ms}^{-1}$ ]
(a) $9.4 \mathrm{~nm}$
(b) $12.3 \mathrm{~nm}$
(c) $10.8 \mathrm{~nm}$
(d) $11.4 \mathrm{~nm}$
Show Answer
Answer:
Correct Answer: 7.( c )
Solution:
- Number of spectral lines produced as an excited electron falls to ground state $(n=1)$ is, $ N=\frac{n(n-1)}{2} $
In given case, $N=6$ $ \begin{aligned} \therefore & & 6=\frac{n(n-1)}{2} \\ \Rightarrow & & n=4 \end{aligned} $
So, $\mathrm{L}^{++}$ electron is in it’s 3rd excited state. Now, using the expression of energy of an electron in $n$th energy level, $ E_n=-\frac{13.6 Z^2}{n^2} \mathrm{eV} $
where, $Z$ is the atomic number. $\therefore$ Energy levels of $\mathrm{L}^{++}$ electron are as shown
$\begin{array}{rlrl} \Delta E & =\frac{h c}{\lambda} \Rightarrow(13.6 \times 9-0.85 \times 9)=\frac{h c}{\lambda} \\ \Rightarrow & \lambda =\frac{h c}{9(13.6-0.85)} \\ \Rightarrow & \lambda =\frac{1240 \mathrm{eV} \cdot \mathrm{nm}}{9 \times 12.75 \mathrm{eV}}=10.8 \mathrm{~nm}\end{array}$
BETA
