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The ionization energy of a H like Bohr's atom is 4 Rydberg.Calculate the wavelength radiated when electron jumps from the first excited state to ground state.

$\begin{array}{1 1}(a)\;203.89A^{\large\circ}&(b)\;303.89A^{\large\circ}\\(c)\;209A^{\large\circ}&(d)\;114A^{\large\circ}\end{array}$

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Energy of I orbit of H like atom =$4R_H$
$\Rightarrow 4\times 2.18\times 10^{-18}$ J
$E_1$ for H=$-2.18\times 10^{-18}J$
$E_H$ like atom =$E_{1H}\times z^2$
$-4\times 2.18\times 10^{-18}=-2.18\times 10^{-18}\times z^2$
(i.e) Atomic no of H like atom is 2 or it is $He^+$ for de-excitation of e in $He^+$ from $n_2=2$ to $n_1=1$
$E_2-E_1=\large\frac{hc}{\lambda}$
$E_1=-4R_h$
$E_2=-\large\frac{4R_h}{4}$$=-R_H$
$E_2-E_1=3R_h=3\times 2.18\times 10^{-18}J$
$\therefore E_2-E_1=\large\frac{hc}{\lambda}$
$\lambda =\large\frac{6.625\times 10^{-34}\times 3.0\times 10^8}{3\times 2.18\times 10^{-18}}$
$\Rightarrow 303.89\times 10^{-10}m$
$\Rightarrow 303.89A^{\large\circ}$
Hence (b) is the correct answer.
answered Jan 17, 2014 by sreemathi.v
 

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