Modern Physics 7 Question 51
57. A gas of identical hydrogen-like atoms has some atoms in the lowest (ground) energy level $A$ and some atoms in a particular upper (excited) energy level $B$ and there are no atoms in any other energy level. The atoms of the gas make the transition to a higher energy level by absorbing monochromatic light of photon energy $2.7 eV$. Subsequently, the atoms emit radiation of only six different energy photons. Some of the emitted photons have an energy of $2.7 eV$, some have more energy and some less than $2.7 eV$.
$(1989,8 M)$
(a) Find the principal quantum number of the initially excited level $B$.
(b) Find the ionization energy for the gas atoms.
(c) Find the maximum and the minimum energies of the emitted photons.
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Answer:
Correct Answer: 57. (a) 2 (b) $14.4 eV$ (c) $13.5 eV, 0.7 eV$
Solution:
- (a) In emission spectrum total six lines are obtained. Hence, after excitation if $n _f$ be the final principal quantum number then,
$$ \frac{n _f\left(n _f-1\right)}{2}=6 \quad \text { or } n _f=4 z $$
i.e. after excitation atom goes to $4^{\text {th }}$ energy state. Hence, $n _i$ can be either 1,2 or 3 .
Absorption and emission spectrum corresponding to $n _i=1, n _i=2$ and $n _i=3$ are shown in figure.
For $n _i=1$, energy of emitted photons $\leq 2.7 eV$
For $n _i=2$, energy of emitted photons $>=<2.7 eV$ and
For $n _i=3$, energy of emitted photons $\geq 2.7 eV$.
As per the given condition $n _i=2$.
(b)
or
$$ E _4-E _2=2.7 eV $$
$$ \frac{E _1}{16}-\frac{E _1}{4}=2.7 $$
or
$$ \begin{aligned} -\frac{3}{16} E _1 & =2.7 \\ E _1 & =-14.4 eV \end{aligned} $$
Therefore, ionization energy for the gas atoms is $\left|E _1\right|$ or $14.4 eV$.
(c) Maximum energy of the emitted photons is corresponding to transition from $n=4$ to $n=1$.
$$ \begin{aligned} \therefore \quad E _{\max } & =E _4-E _1 \\ & =\frac{E _1}{16}-E _1=\frac{-15 E _1}{16} \\ & =-\frac{15}{16}(14.4)=13.5 eV \end{aligned} $$
Similarly, minimum energy of the emitted photons is corresponding to transition from $n=4$ to $n=3$.
$$ \begin{aligned} \therefore \quad E _{\min } & =E _4-E _3 \\ & =\frac{E _1}{16}-\frac{E _1}{9}=-\frac{7 E _1}{16 \times 9} \\ & =\frac{7 \times 14.4}{16 \times 9}=0.7 eV \end{aligned} $$
