Question

If radius of the $1^{st}$ orbit of $H$-atom is $0.0529 \,nm$ then the radius of its $3^{rd}$ orbit is

• A. $0.05328 \,nm$
• B. $0.05121\, nm$
• C. $0.04121 \,nm$
• D. $0.00587 \,nm$

Answer 1

Answer: (D)

$0.00587 \,nm$

Answer 2

Radius of $n^{th}$ orbit in hydrogen atom is $r_{n}=\frac{r_{1}}{n^{2}}$ where $r_{1}$ is the radius of $1^{st}$ orbit. Here, $r_{1} = 0.0529 \,nm$ For $3^{rd}$ orbit, $n=3$, $\therefore r_{3}=\frac{0.0529\,nm}{3^{2}}$ $=0.00587\,nm$