Question

As the orbit number increases, the distance between two consecutive orbits (11 = radius of first orbit)
(A) increases by 2r;
(B) increases by (2n-1), where n is lower orbit number
(C) increases by (2n-1), where n is higher orbit number
(D) remains constant​

Answer 1

Hey Dear,

◆ Answer -

(C) ∆r = r1.(2n-1) where n = upper orbit no

◆ Explanation -

Radius of Bohr's orbit is directly proportional to square of orbit number.

For orbit no n,

r = r1.n²

For lower orbit,

r' = r1.(n-1)²

r' = r1.(n²-2n+1)

Distance between consecutive orbits is -

∆r = r - r'

∆r = r1.n² - r1.(n²-2n+1)

∆r = r1.(n²-n²+2n-1)

∆r = r1.(2n-1)

Hence, distance between two consecutive orbit radius is r1(2n-1).

Thanks dear...