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Q. An electron is moving in an orbit of a hydrogen atom from which there can be a maximum of six transitions. An electron is moving in an orbit of another hydrogen atom from which there can be a maximum of three transitions. The ratio of the velocity of the electron in these two orbits is

KCETKCET 2010Atoms

Solution:

Number of spectral lines obtained due to transition of electrons from $n$ th orbit to lower orbit is,
$N=\frac{n(n-1)}{2}$
I case $6=\frac{n_{1}\left(n_{1}-1\right)}{2}$
$\Rightarrow n_{1} =4$
II case $3=\frac{n_{2}\left(n_{2}-1\right)}{2}$
$\Rightarrow n_{2}=3$
Velocity of electron in hydrogen atom in $n$ th orbit
$v_{n} \propto \frac{1}{n}$
$\frac{v_{n}}{v_{n}'} =\frac{n_{2}}{n_{1}}$
$\Rightarrow \frac{v_{6}}{v_{3}} =\frac{3}{4}$