JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 11)
The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 $$ \times $$ 10-16 s. The frequency of revolution of the electron in its first excited state (in s-1) is :
5.6 $$ \times $$ 1012
1.6 $$ \times $$ 1014
7.8 $$ \times $$ 1014
6.2 $$ \times $$ 1015
Explanation
Time period of revolution of electron in nth orbit
V = $${{2\pi r} \over V}$$
= $${{2\pi {a_0}\left( {{{{n^2}} \over Z}} \right)} \over {{V_0}\left( {{Z \over n}} \right)}}$$
$$ \Rightarrow $$ T $$ \propto $$ $${{{{n^3}} \over {{Z^2}}}}$$
$$ \therefore $$ $${{{T_1}} \over {{T_2}}} = {{n_1^3} \over {n_2^3}}$$
$$ \Rightarrow $$ $${{1.6 \times {{10}^{ - 16}}} \over {{T_2}}} = {1 \over {{{\left( 2 \right)}^3}}}$$
$$ \Rightarrow $$ T2 = 1.6 $$ \times $$ 8 $$ \times $$ 10-16
$$ \therefore $$ f2 = $${1 \over {{T_2}}}$$ = $${1 \over {1.6 \times 8 \times {{10}^{ - 16}}}}$$ = 7.8 $$ \times $$ 1014
V = $${{2\pi r} \over V}$$
= $${{2\pi {a_0}\left( {{{{n^2}} \over Z}} \right)} \over {{V_0}\left( {{Z \over n}} \right)}}$$
$$ \Rightarrow $$ T $$ \propto $$ $${{{{n^3}} \over {{Z^2}}}}$$
$$ \therefore $$ $${{{T_1}} \over {{T_2}}} = {{n_1^3} \over {n_2^3}}$$
$$ \Rightarrow $$ $${{1.6 \times {{10}^{ - 16}}} \over {{T_2}}} = {1 \over {{{\left( 2 \right)}^3}}}$$
$$ \Rightarrow $$ T2 = 1.6 $$ \times $$ 8 $$ \times $$ 10-16
$$ \therefore $$ f2 = $${1 \over {{T_2}}}$$ = $${1 \over {1.6 \times 8 \times {{10}^{ - 16}}}}$$ = 7.8 $$ \times $$ 1014
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