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A single electron orbits around a stationary nucleus of charge +ze , where z is a constant and e is magnitude of electronic charge. It requires 47.2eV to exite the electron from $\;2^{nd}\;$ Bohr orbital to $\;3^{rd}\;$ Bohr orbital. find the atomic no.of element.


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Answer : (c) 5
Explanation :
Position is $\;n_{1}=2\; ----> \;n_{2}=3\;:\bigtriangleup \varepsilon=47.2eV$
We have $\bigtriangleup \varepsilon=13.6\;z^2\;(\large\frac{1}{n_{1}^{2}}-\large\frac{1}{n_{2}^{2}})eV$
answered Feb 22, 2014 by yamini.v

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