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The momentum of an electron revolving in \mathrm{n}^{\text {th }} orbit is given by : (Symbols have their usual meanings)
 

Option: 1

\frac{\mathrm{nh}}{2 \pi \mathrm{r}}


Option: 2

\frac{\mathrm{nh}}{2 \mathrm{r}}


Option: 3

\frac{\mathrm{nh}}{2 \pi}


Option: 4

\frac{2 \pi r}{n h}


Answers (1)

best_answer

The angular momentum of an electron in \mathrm{n^{\text {th }}} orbit is.

\mathrm{L_n=m v_n r_n=\frac{n h}{2 \pi} }

\mathrm{P_n=m v_n=\frac{n h}{2 \pi r_n} }

\mathrm{P_n=\frac{n h}{2 \pi r}}
Hence (1) is correct option

Posted by

sudhir.kumar

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