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

Question

Difference between n text th and (n+1) text th Bohr's radius of H atom is equal to it's ( n -1) text th Bohr's radius. The value of n is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

( n +1)2 a 0- n 2 a 0=( n -1)2 a 0 n 2+2 n +1- n 2=( n -1)2 2 n +1= n 2-2 n +1 4 n = n 2 n =4

Q. 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

1

2

3

4

$(n + 1)^{2} a_{0} - n^{2} a_{0} = (n - 1)^{2} a_{0}$
$n^{2} + 2 n + 1 - n^{2} = (n - 1)^{2}$
$2 n + 1 = n^{2} - 2 n + 1$
$4 n = n^{2}$
$n = 4$

Q. Difference between nth 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

1

2

3

4

(n+1)2a0​−n2a0​=(n−1)2a0​ n2+2n+1−n2=(n−1)2 2n+1=n2−2n+1 4n=n2 n=4

Q. 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

$(n + 1)^{2} a_{0} - n^{2} a_{0} = (n - 1)^{2} a_{0}$
$n^{2} + 2 n + 1 - n^{2} = (n - 1)^{2}$
$2 n + 1 = n^{2} - 2 n + 1$
$4 n = n^{2}$
$n = 4$