Question

Which one is likely to be the orbit number for a circular orbit of diameter $8940.10\text{pm}$ of the hydroen atom, if we assume Bohr orbit to be the same as that represented by the principal quantum number?

• A. $14$
• B. $13$
• C. $12$
• D. $16$

Answer 1

The correct option is B $13$

According to the question,
radius of Bohr's Orbit = $8940.10\text{pm}$
we know,
$\text{Radius}\left(r\right)=0.529\frac{n^2}{Z}{\text{A}}^{\circ }$
For hydrogen atom, $Z=1$
$\therefore 89.401=0.529n^2$
$\Rightarrow n^2=\frac{89.401}{0.529}=169$
$n=13$