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 approximately ___.

Answer 1

We know that

$r_n\propto n^2$

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

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

$n=4$