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A small particle of mass $m$ moves in such a way that the potential energy of particle is given as $\mathrm{U}=-\frac{1}{2} \mathrm{~m} \alpha^2 \mathrm{r}^2$ where $\alpha$ is constant and $\mathrm{r}$ is the distance of particle from centre. If Bohr's model of quantization of angular momentum and circular orbit is valid for the particle. ( $\mathrm{h}=$ Planck's constant). Total energy of particle in its orbit is
Option: 1
$-\frac{n h \alpha}{4 \pi}$

Option: 2
$\frac{\mathrm{nh} \alpha}{2 \pi}$

Option: 3
$zero$

Option: 4
$-\frac{\mathrm{nh} \alpha}{2 \pi}$

Answers (1)

best_answer

$\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dr}}=\frac{\mathrm{mV}_{\mathrm{n}}^2}{\mathrm{r}_{\mathrm{n}}}$

$m \alpha^2 r_n=\frac{m V_n^2}{r_n}$ $(1)$

$\text { and } \mathrm{mv}_{\mathrm{n}} \mathrm{r}_{\mathrm{n}}=\frac{\mathrm{nh}}{2 \pi}$ $(2)$

Solving equations (1) and (2)

$\mathrm{r}_{\mathrm{n}}=\left(\frac{\mathrm{nh}}{2 \pi \mathrm{m \alpha}}\right)^{1 / 2}$

$\text { and } \mathrm{K}=\frac{1}{2} \mathrm{mv}_{\mathrm{n}}^2=\frac{\mathrm{nh} \alpha}{4 \pi}$

$\text { Total energy } \mathrm{E}=\mathrm{U}+\mathrm{K}=-\frac{1}{2} \mathrm{~m} \alpha^2 \mathrm{r}_{\mathrm{n}}^2+\frac{1}{2} \mathrm{mV}_{\mathrm{n}}^2=0$

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shivangi.bhatnagar

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