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A small particle of mass m moves in such a way that its potential energy   \mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{r}^2
is constant and r is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit,
the radius of nth orbit will be proportional to,

Option: 1

n


Option: 2

n^{2}


Option: 3

\frac{1}{n}


Option: 4

\sqrt{n}


Answers (1)

best_answer

Given  \mathrm{U}=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{r}^2   to find radius r as f (n), where n is orbit

\\ |\frac{1}{2} \times \mathrm{P} \cdot \mathrm{E}|=|\mathrm{K} \cdot \mathrm{E} |\\ \\ \frac{1}{2}\times \frac{1}{2} \mathrm{~m} \omega^2 \mathrm{r}^2 =\frac{1}{2}\times mv^2 \\ \\ v \ \ \alpha \ \ \omega r

Using Bohr’s postulate: angular momentum L = mvr =\frac{nh}{2\pi}

or   \mathrm{mr^2} \omega \ \alpha \ \ \frac{\mathrm{nh}}{2 \pi}

\Rightarrow r \propto \sqrt{n}

 

 

Posted by

Divya Prakash Singh

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