An electron in hydrogen occupies its second Bohr orbit. An electron in the hydrogen-like ion Be³⁺ is in an orbit having exactly the same radius. If the principal quantum number of the Be³⁺ electron is N, find N.
Physics · Atomic Structure and X-rays · atomic structure and bohr atomic model
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Question

An electron in hydrogen occupies its second Bohr orbit. An electron in the hydrogen-like ion Be³⁺ is in an orbit having exactly the same radius. If the principal quantum number of the Be³⁺ electron is N, find N.

Physics · Atomic Structure and X-rays · atomic structure and bohr atomic model

G1 Physics · Accepted Answer

For hydrogen in the second orbit: r(H) = a₀×2² ÷ 1 = 4a₀. For Be³⁺, nuclear charge Z = 4: r(Be³⁺) = a₀N² ÷ 4. The radii are equal: N²a₀ ÷ 4 = 4a₀. Cancel a₀: N² ÷ 4 = 4. Thus, N² = 16, N = 4.