Question

The time period of revolution in the third orbit of Li^2+ion is a x second The time period of revolution in the second orbit of He+ion should be

Answer 1

Answer:

        3 / 2

Explanation:

For Li²⁺ , n = 3

               z = 3

               T = T₁

For He⁺ , n = 2

               z = 2

               T = T₂

T ∝ n³ / z²

T₁ = K( 3³ / 3² ) ----{i}

T₂ = K( 2³ / 2² ) ----{ii}

{i} / {ii}

T₁ / T₂ = K( 3³ / 3² ) ÷ K( 2³ / 2² )

          = 3³ / 3²÷  3³ / 2²

          =  3³ / 3² ×  2² / 2³

          = 3 × 1/2

          = 3 / 2