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In accordance with Bohr's model, find the quantum number that characterises the earth's revolution around the sun in an orbit of radius 1.5×10¹¹ m with orbital speed 3×10⁴ m s⁻¹. (Mass of earth =6×10²⁴ kg, h=6.625×10⁻³⁴ J s)

Asked in GUJCET 2022 · Bohr's rules in other systems

• (1) 3.6×10⁷⁴
• (2) 1.6×10⁷⁴
• (3) 2.6×10⁷⁴
• (4) 4.6×10⁷⁴

Step-by-step solution

Why the other options are wrong

• (1) No single slip gives 3.6×10⁷⁴. Sanity check: n=(2π mvr)/h=(2π×2.7×10⁴⁰)/(6.625×10⁻³⁴)=2.56×10⁷⁴.
• (2) No single slip gives 1.6×10⁷⁴. Sanity check: leaving out the 2π would give 4.1×10⁷³, and the full expression gives 2.56×10⁷⁴.
• (4) No single slip gives 4.6×10⁷⁴; it differs from the answer only in the leading digit. Sanity check: mvr=2.7×10⁴⁰ kg m² s⁻¹, so n=(2π mvr)/h≈2.6×10⁷⁴.

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