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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 . Find the value of n for which radius of orbit is approximately the same as that of first Bohr orbit for hydrogen atom.


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Answer : (b) 25
Explanation :
The radius of first Bohr orbit for hydrogen is :
$r_{1}=\large\frac{\varepsilon_{0}h^2}{\pi m_{e}e^2}$
For $\;r_{n}\;$ ($\;mu\;$-mesonic atom) = $\;r_{1}\;$ (hydrogen atom) ,
We have
$\large\frac{\varepsilon_{0} n^2 h^2}{624 \pi m_{e} e^2}=\large\frac{\varepsilon_{0} h^2}{n m_{e} e^2}$
$n^2=624 \qquad \; n=25\;.$
answered Feb 26, 2014 by yamini.v

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