1. r ∝ n^(1/3);\ v ∝ n^(1/3) 2. r ∝ n^(1/3);\ v ∝ n^(2/3) 3. r ∝ n^(2/3);\ v ∝ n^(1/3) 4. r ∝ n^(4/3);\ v ∝ n^(-1/3)

Answer: 3. r ∝ n^(2/3);\ v ∝ n^(1/3) Key idea: apply Newton's second law with the constant force as the centripetal force, together with Bohr's quantization mvr=nhbar . mv²/r=F,qquad mvr=nh/2π From the second, v=nh/2π mr . Substituting into the first: m/r·n²h²/4π²m²r²=F⇒ r³∝ n²⇒ r∝ n^(2/3) Then v∝n/r∝ n^(1-2/3)=n^(1/3) .

Topic: Bohr model, energy levels, hydrogen spectrum· Atoms and Nuclei

A particle of mass m is moving around the origin with a constant force F pulling it towards the origin. If Bohr model is used to describe its motion, the radius r of the n^th orbit and the particle's speed v in the orbit depend on n as

1. $r \propto n^{1/3};\ v \propto n^{1/3}$

2. $r \propto n^{1/3};\ v \propto n^{2/3}$

3. $r \propto n^{2/3};\ v \propto n^{1/3}$

4. $r \propto n^{4/3};\ v \propto n^{-1/3}$

Doubtly Asked 30 Oct 2026 NEET UG 2025 Practice test neet Save WhatsApp Report

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Answer: 3. $r \propto n^{2/3};\ v \propto n^{1/3}$

Key idea: apply Newton's second law with the constant force as the centripetal force, together with Bohr's quantization $mvr=n\hbar$.

$$\frac{mv^2}{r}=F,\qquad mvr=\frac{nh}{2\pi}$$

From the second, $v=\dfrac{nh}{2\pi mr}$. Substituting into the first:

$$\frac{m}{r}\cdot\frac{n^2h^2}{4\pi^2m^2r^2}=F\Rightarrow r^3\propto n^2\Rightarrow r\propto n^{2/3}$$

Then $v\propto\dfrac nr\propto n^{1-2/3}=n^{1/3}$.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A particle of mass m is moving around the origin with a constant force F pulling it towards the origin. If Bohr model is used to describe its motion, the radius r of the n^th orbit and the particle's speed v in the orbit depend on n as

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