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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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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.

$(a)\;1\qquad(b)\;3\qquad(c)\;5\qquad(d)\;7$

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

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