Modern Physics Ques 64
- A doubly ionised lithium atom is hydrogen-like with atomic number $3$.
$(1985,6 M)$
(a) Find the wavelength of the radiation required to excite the electron in $Li^{2+}$ from the first to the third Bohr orbit. (Ionisation energy of the hydrogen atom equals $13.6$ $ eV$.)
(b) How many spectral lines are observed in the emission spectrum of the above excited system?
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Answer:
Correct Answer: 64.(a) $ 113.74 $ $\AA \quad $ (b) $3$
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
- Given $Z=3, E _n \propto \frac{Z^{2}}{n^{2}}$
(a) To excite the atom from $n=1$ to $n=3$, energy of photon required is
$ \begin{aligned} E _{1-3}=E _3-E _1 & =\frac{(-13.6)(3)^{2}}{(3)^{2}}\left[-\frac{(-13.6)(3)^{2}}{(1)^{2}}\right] \\ & =108.8 eV \end{aligned} $
Corresponding wavelength will be,
$ \lambda(\text { in } \AA)=\frac{12375}{E(\text { in eV })}=\frac{12375}{108.8}=113.74 $ $\AA $
(b) From $n^{\text {th }}$ orbit total number of emission lines can be $\frac{n(n-1)}{2}$.
$\therefore \quad $ Number of emission lines $=\frac{3(3-1)}{2}=3$
BETA
