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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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A particle of charge equal to that of an electron e , and mass 208 times the mass of electron (called $\;\mu-\;$ meson ) moves in a circular orbit around a nucleus of charge +3e . (Take the mass of nucleus to be infinite ) . Then radius of $n^{th}\;$ Bohr orbit is

$(a)\;\large\frac{\varepsilon_{0} n^2 h^2}{2 \pi m_{e} e^2}\qquad(b)\;\large\frac{\varepsilon_{0} n^2 h^2}{4 \pi m_{e}e^2} \qquad(c)\;\large\frac{\varepsilon_{0} n^2 h^2}{8 \pi m_{e}e^2} \qquad(d)\;\large\frac{\varepsilon_{0} n^2 h^2}{624 \pi m_{e}e^2} $

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Answer : $\;\large\frac{\varepsilon_{0} n^2 h^2}{624 \pi m_{e}e^2} $
Explanation :
We have radius of $\;n^{th}\;$ orbit given by
$r_{n}=\large\frac{n^2 h^2 \varepsilon_{0}}{\pi m e^2 z}$
$K=\large\frac{1}{4 \pi \varepsilon_{0}}\;,Z=3 \;,$ & m=208 m_{e}
We get
$r_{n}=\large\frac{\varepsilon_{0} n^2 h^2}{624 \pi m_{e} e^{2}}\;.$
answered Feb 26, 2014 by yamini.v
 

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