Question

If the radius of first Bohr orbit be $a_0$, then the radius of the third orbit would be -

• A. $3\times a_o$
• B. $6\times a_o$
• C. $9\times a_o$
• D. $1/9\times a_o$

Answer 1

Answer: (C)

$r=0.529\frac{n^2}{Z}\mathring{A}$
$n=1$,
$Z=1$
so $r=0.529\mathring{A}=a_o$
Therefore for third orbit
$r=0.529\times 9$
$orr=9\times a_o$