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Q.
In the following transitions, which one has higher frequency?
ManipalManipal 2010Atoms
Solution:
From Bohr's postulate, energy of electron in $n t h$ orbit is given by
$E=-\frac{M Z^{2} e^{4}}{8 \varepsilon_{0}^{2} h^{2}}\left(\frac{1}{n^{2}}\right)$
When electron jumps from some higher energy state $n_{2}$ to a lower energy state $n_{1}$, the energy difference between these states is
$E_{2}-E_{1} \propto\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
From Bohr's third postulate, the frequency $v$ of the electromagnetic wave is
$v=\frac{E_{2}-E_{1}}{h} \propto\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
First case
$ n_{1}=1, n_{2}=3$
$\therefore v_{1} \propto\left(1-\frac{1}{9}\right) \propto \frac{8}{9}$
Second case $n_{1}=2, n_{2}=4$
$\therefore v_{2} \propto\left(\frac{1}{4}-\frac{1}{16}\right) \propto \frac{3}{16}$
Third case $ n_{1}=3, n_{2}=4 $
$\therefore v_{3} \propto\left(\frac{1}{9}-\frac{1}{16}\right) \propto \frac{7}{144}$
Fourth case $ n_{1}=2, n_{2}=3$
$\therefore v_{4} \propto\left(\frac{1}{4}-\frac{1}{9}\right) \propto \frac{5}{36} $
$ v_{1}>v_{2}>v_{4}>v_{3}$
Hence, transition $3 \rightarrow 1$ has higher frequency.