Question

Difference between $n^{th}$ and $(n+1{)}^{th}$ Bohr's radius of H atom is equal to it's $(n-1{)}^{th}$ Bohr's radius. The value of n is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

We know that $r_n\propto n^2$

Given: $r_{n+1}-r_n=r_{n-1}$

$\therefore (n+1{)}^2-n^2=(n-1{)}^2$

$n=4$