Question

If an electron is revolving in its Bohr orbit having Bohr radius of 0.529 Å, then the radius of third orbit is

• A. 4234 nm
• B. 4496 Å
• C. 4.71 Å
• D. 5125 nm

Answer 1

Answer: (C)

4.71 Å

Answer 2

As $r_{n} = n^{2}a_{0}$

Here $a_{0} = 0.529Åandn = 3$

$\therefore r_{n = 3} = (3)^{2}a_{0}$

$= 9 \times 0.529Å = 4.761Å$