JEE MAIN - Chemistry (2017 - 9th April Morning Slot - No. 21)
The electron in the hydrogen atom undergoes transition from higher orbitals
to orbital of radius 211.6 pm. This transition is associated with :
Lyman series
Balmer series
Paschen series
Brackett series
Explanation
Here electron is coming to the orbital of radius 211.6 pm. Now we have to find which series have radius of orbital 211.6 pm.
We know,
Radius, r = $$0.529 \times {{{n^2}} \over Z}\,\mathop A\limits^ \circ $$
Given, r = 211.6 pm = 211.6 $$ \times $$ 10-12 m
and Z = 1 for hydrogen atom
$$ \therefore $$ 211.6 $$ \times $$ 10-12 = $$0.529 \times {{{n^2}} \over 1}$$ $$ \times $$ 10-10
$$ \Rightarrow $$ n = 2
As n = 2 so the series is Balmer Series.
We know,
Radius, r = $$0.529 \times {{{n^2}} \over Z}\,\mathop A\limits^ \circ $$
Given, r = 211.6 pm = 211.6 $$ \times $$ 10-12 m
and Z = 1 for hydrogen atom
$$ \therefore $$ 211.6 $$ \times $$ 10-12 = $$0.529 \times {{{n^2}} \over 1}$$ $$ \times $$ 10-10
$$ \Rightarrow $$ n = 2
As n = 2 so the series is Balmer Series.
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