Two electrons are revolving in orbits of two different hydrogen-like atoms. The time periods of revolution of the electrons in these orbits are equal. If their speeds are 2 10^6 m/s and 3 10^6 m/s respectively, the possible order of their principal quantum numbers is:

Step-by-step solution

For hydrogen-like atoms, the time period of revolution is T n^3 Z^2 and the speed is v Z n . From the speed relation, we have Z n v . Substituting this into the time period relation gives: T n^3 (n v)^2 = n v^2 Since the time periods are equal ( T₁ = T₂ ): n₁ v₁^2 = n₂ v₂^2 n₁ n₂ = ( v₁ v₂ )^2 Given v₁ = 2 10^6 m/s and v₂ = 3 10^6 m/s : n₁ n₂ = ( 2 10^6 3 10^6 )^2 = 4 9 The possible order of their principal quantum numbers is 4 and 9 . Answer: 4 and 9