Modern Physics Ques 62
- A hydrogen like atom (atomic number $Z$ ) is in a higher excited state of quantum number $n$. The excited atom can make a transition to the first excited state by successively emitting two photons of energy $10.2 $ $eV$ and $17.0 $ $eV$ respectively. Alternately, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies $4.25$ $ eV$ and $5.95$ $ eV$ respectively.
$(1994,6 M)$
Determine the values of $n$ and $Z$.
(Ionization energy of $H$-atom $=13.6 $ $eV$ )
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Answer:
Correct Answer: 62.$(6,3)$
Solution:
Formula:
- From the given conditions
$ \begin{aligned} & E _n-E _2=(10.2+17) eV=27.2 eV \quad …….(i)\\ & \text { and } \quad E _n-E _3=(4.25+5.95) eV=10.2 eV\quad …….(ii) \end{aligned} $
Eq. (i) - Eq. (ii) gives
$ \begin{aligned} E _3-E _2 & =17.0 eV \\ \text { or } \quad Z^{2}(13.6) (\frac{1}{4}-\frac{1}{9}) & =17.0 \end{aligned} $
$ \begin{array}{ll} \Rightarrow & Z^{2}(13.6)(5 / 36)=17.0 \\ \Rightarrow & Z^{2}=9 \text { or } Z=3 \end{array} $
From Eq. (i) $Z^{2}$ (13.6) $(\frac{1}{4}-\frac{1}{n^{2}})=27.2$
$ \begin{aligned} & \text { or } \quad(3)^{2}(13.6) (\frac{1}{4}-\frac{1}{n^{2}})=27.2 \\ & \text { or } \quad \frac{1}{4}-\frac{1}{n^{2}}=0.222 \\ & \text { or } \quad 1 / n^{2}=0.0278 \text { or } n^{2}=36 \\ & \therefore \quad n=6 \end{aligned} $
BETA
