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A particle of mass m moves in a circular orbit in a central potential field U(r) = $\frac{1}{2} k r^{2}$ . If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as :
Option 1)
$r_{n} \ \alpha \ \sqrt{n} , E_{n} \ \alpha \ \frac{1}{n}$
Option 2)
$r_{n} \ \alpha \ n , E_{n} \ \alpha \ n$
Option 3)
$r_{n} \ \alpha \ \sqrt{n} , E_{n} \ \alpha \ n$
Option 4)
$r_{n} \ \alpha \ n^{2} , E_{n} \ \alpha \ \frac{1}{n^{2}}$

Answers (1)

best_answer

Hypothetical atomic model -

$\left | \frac{dU}{dr} \right |= \frac{mv^{2}}{r}$

$mvr= \frac{nh}{2\pi }$

- wherein

Solve these two equations to get radius and energy of hypothetical atoms

$F=\frac{-du}{dr}$

$U=\frac{1}{2}kr^{2}$

$F=\frac{1}{2}\times 2\times kr=kr$

$F=\frac{mv^{2}}{r}=kr$

So $v^{2}\alpha\, r^{2}$

$v\alpha\, r$

& form both quantisation principle

$mv r=\frac{nh}{2\pi }$

So $vr\, \, \alpha\, \, n$

$\Rightarrow r^{2}\,\alpha \, n$

$\Rightarrow r\,\alpha \, \sqrt{n}$

now $U =E=\frac{1}{2}kr^{2}$

$U \alpha \, r^{2}$

$U \alpha \, n$

$E \alpha \, n$

Option 1)

$r_{n} \ \alpha \ \sqrt{n} , E_{n} \ \alpha \ \frac{1}{n}$

Option 2)

$r_{n} \ \alpha \ n , E_{n} \ \alpha \ n$

Option 3)
$r_{n} \ \alpha \ \sqrt{n} , E_{n} \ \alpha \ n$
Option 4)

$r_{n} \ \alpha \ n^{2} , E_{n} \ \alpha \ \frac{1}{n^{2}}$

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