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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.The wavelength of radiation required remove electron from first Bohr's orbit to infinity is

$(a)\;39.2 A^{0}\qquad(b)\;36.6 A^{0}\qquad(c)\;42.1 A^{0}\qquad(d)\;31.2 A^{0}$

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Answer : (b) $\;36.6 A^{0}$
Explanation :
Position is $\;n_{1}=1 \; ---> y\quad n_{2}=\infty$
$\bigtriangleup \varepsilon=ionization\;energy=?$
$\bigtriangleup \varepsilon=13.6\times5^2\;(\large\frac{1}{1^2}-\large\frac{1}{\infty^2})=340 eV$
$\lambda=\large\frac{hC}{\bigtriangleup \varepsilon}=\large\frac{6.63\times10^{-34}\times3\times10^{8}}{340\times1.6\times10^{-19}}=36.56\times10^{-10} A^{0}\;.$
answered Feb 22, 2014 by yamini.v
 

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