The time period of revolution of an electron in the first stationary orbit of a He^+ ion is T . The time period of revolution of the electron in the second stationary orbit of a Li²⁺ ion will be

Step-by-step solution

The time period of revolution of an electron in the n^ th Bohr orbit is given by T = 2 r v . Since the radius r n^2 Z and the speed v Z n , the time period T n^3 Z^2 . For the first stationary orbit of He^+ ( n = 1 , Z = 2 ): T₁ 1^3 2^2 = 1 4 For the second stationary orbit of Li²⁺ ( n = 2 , Z = 3 ): T₂ 2^3 3^2 = 8 9 Taking the ratio of the time periods: T₂ T₁ = 8/9 1/4 = 32 9 Therefore, T₂ = 32 9 T₁ = 32 9 T . Answer: 32 9 T