Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 × 10⁵ m/s and 2.5 × 10⁵ m/s respectively. If the radii of these orbits are nearly same then the possible order of energy states ar $e_____$ respectively.
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Question

Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 × 10⁵ m/s and 2.5 × 10⁵ m/s respectively. If the radii of these orbits are nearly same then the possible order of energy states ar $e_____$ respectively.

Options

G1 Physics · Accepted Answer

For a hydrogen-like atom, v ∝ Z/n and r ∝ n²/Z. Eliminate Z. From v ∝ Z/n, we get Z ∝ vn. Therefore, r ∝ n²/(vn) = n/v. The two orbital radii are nearly the same, so n₁/v₁ = n₂/v₂. Hence, n₁/n₂ = v₁/v₂ = (3×10⁵)/(2.5×10⁵) = 6/5. Therefore a possible pair of integral quantum numbers is n₁ = 6, n₂ = 5. Correct option: A.

Two electrons are moving in orbits of two hydrogen like atoms with speeds 3 × 10⁵ m/s and 2.5 × 10⁵ m/s respectively. If the radii of these orbits are nearly same then the possible order of energy states ar e_____ respectively.