Modern Physics 1 Question 50
46. 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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Solution:
- (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.
$$ \begin{aligned} E _m & =-0.85 eV \\ \text { or } \quad-13.6 \frac{z^{2}}{m^{2}} & =-0.85 \\ \text { or } \quad \frac{z}{m} & =0.25 \end{aligned} $$
Similarly, $\quad E _{m+3}=-0.544 eV$
or $-13.6 \frac{z^{2}}{(m+3)^{2}}=-0.544$
or $\quad \frac{z}{(m+3)}=0.2$
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 $E _{m+3}-E _m$
$$ \begin{aligned} \therefore \quad \lambda _{\min } & =\frac{h c}{\Delta E _{\max }} \\ & =\frac{1240}{0.306}=4052.3 nm . \end{aligned} $$
$$ \therefore \quad \Delta E _{\max }=-0.544-(-0.85)=0.306 eV $$
