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Calculate $\lambda$ of the radiations when the electrons jumps from III to II orbit for H atom.The electronic energy in II and III Bohr's orbit of H atom are $-5.42\times 10^{-12}$ and $-2.41\times 10^{-12}$erg respectively.


1 Answer

$E_3$ for H=$-2.41\times 10^{-21}erg$
$E_2$ for H=$-5.42\times 10^{-21}erg$
$\therefore$ For a jump from III to II shell
$\Delta E=E_3-E_2=\large\frac{hc}{\lambda}$
$\therefore \lambda=\large\frac{hc}{E_3-E_2}$
$\Rightarrow \large\frac{6.625\times 10^{-27}\times 3.0\times 10^{10}}{-2.41\times 10^{-12}+5.42\times 10^{-12}}$
$\Rightarrow 6602.9\times 10^{-8}$cm
$\Rightarrow 6602.9A^{\circ}$
Hence (a) is the correct answer.
answered Jan 16, 2014 by sreemathi.v

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