JEE MAIN - Chemistry (2020 - 8th January Evening Slot - No. 7)
The radius of the second Bohr orbit, in terms
of the Bohr radius, a0, in Li2+ is :
$${{2{a_0}} \over 9}$$
$${{2{a_0}} \over 3}$$
$${{4{a_0}} \over 9}$$
$${{4{a_0}} \over 3}$$
Explanation
$${r_n} = {{{n^2}{a_0}} \over Z}$$
For 2nd Bohr orbit of Li+2
n = 2
and Z = 3
r = $${{{2^2}{a_0}} \over 3}$$ = $${{4{a_0}} \over 3}$$
For 2nd Bohr orbit of Li+2
n = 2
and Z = 3
r = $${{{2^2}{a_0}} \over 3}$$ = $${{4{a_0}} \over 3}$$
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