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Q.
Which transition emits light of lower frequency ?
AMUAMU 2001
Solution:
From Bohr's concept, the wavelength $(\lambda)$ of emitted light is given by
$\frac{1}{\lambda}=R\left(\frac{1}{n_{1}{ }^{2}}-\frac{1}{n_{2}{ }^{2}}\right)$
For $n_{1}=1,\,\, n_{2}=2$
$\therefore \frac{1}{\lambda_{1}} =R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$
$=R\left(\frac{1}{1}-\frac{1}{4}\right)=0.75\, R$
For $n_{1}=2,\,\, n_{2}=6$
$\frac{1}{\lambda_{2}}=R\left(\frac{1}{4}-\frac{1}{36}\right)=0.22\, R$
Also frequency $(v) \propto \frac{1}{\lambda}$.
and $E=hv$
Hence, for minimum $E, n_{2}=6$ and $n_{1}=2$.