Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 × 10^ 5 ~m / s and 2.5 × 10^ 5 ~m / s respectively. If the radii of these orbits are nearly same then the possible order of energy states are

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Correct answer: D. 6 and 5

For hydrogen-like atoms: V (Z)/(n) and r (n^2)/(Z) Thus r (n)/(V) Since the radii are nearly the same: (n_1)/(V_1) = (n_2)/(V_2) (n_1)/(n_2) = (V_1)/(V_2) = (3 × 10^5)/(2.5 × 10^5) = (6)/(5) Possible order of energy states is 6 and 5.

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