Modern Physics Ques 59
- A hydrogen-like atom (described by the Bohrs model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between $-0.85$ $ eV$ and $-0.544 $ $eV$ (including both these values).
$(2002,5$ M)
(a) Find the atomic number of the atom.
(b) Calculate the smallest wavelength emitted in these transitions.
(Take $h c=1240 $ $eV-$ $nm$, ground state energy of hydrogen atom $=-13.6 $ $eV)$
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Answer:
Correct Answer: 59.(a) $z=3$ (b) $ 4052.3 $ $nm $
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
- (a) Total $6$ lines are emitted. Therefore,
$ \frac{n(n-1)}{2}=6 \quad \text { or } \quad n=4 $
So, transition is taking place between $m^{\text {th }}$ energy state and $(m+3)^{\text {th }}$ energy state.
$ \begin{aligned} E _m & =-0.85 eV \\ \text { or } \quad-13.6 \frac{z^{2}}{m^{2}} & =-0.85 \\ \text { or } \quad \frac{z}{m} & =0.25 \quad …….(i) \end{aligned} $
Similarly, $\quad E _{m+3}=-0.544 $ $eV$
or $-13.6 (\frac{z^{2}}{(m+3))^{2}}=-0.544$
or $\quad \frac{z}{(m+3)}=0.2 \quad …….(ii)$
Solving Eqs. (i) and (ii) for $z$ and $m$, we get
$ m=12 \text { and } z=3 $
(b) Smallest wavelength corresponds to maximum difference of energies which is obviously $E _{m+3}-E _m$
$ \begin{aligned} \therefore \quad \lambda _{\min } & =\frac{h c}{\Delta E _{\max }} \\ & =\frac{1240}{0.306}=4052.3 nm . \end{aligned} $
BETA
