The radius of the second Bohr orbit, in terms of the Bohr radius, a0, in Li2+ is :
Solution
${r_n} = {{{n^2}{a_0}} \over Z}$
<br><br>For 2<sup>nd</sup> Bohr orbit of Li<sup>+2</sup>
<br><br>n = 2
<br><br>and Z = 3
<br><br>r = ${{{2^2}{a_0}} \over 3}$ = ${{4{a_0}} \over 3}$
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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