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Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Atomic Structure

The energy E for an electron in H atom is $\large\frac{21.7\times 10^{-12}}{h^2}$erg.Calculate the energy required to remove electron completely from $n=2$ orbit.Also calculate the longest wavelength of light that can be used to cause this transition.

$\begin{array}{1 1}(a)\;4367.4A^{\large\circ}&(b)\;3263A^{\large\circ}\\(c)\;3663.6A^{\large\circ}&(d)\;2348.1A^{\large\circ}\end{array}$

1 Answer

$E_n=-\large\frac{21.7\times 10^{-12}}{n^2}$erg
$\therefore E_2=-\large\frac{21.7\times 10^{-12}}{4}$$=5.425\times 10^{-12}erg$
For removal of electron $E_2=\large\frac{hc}{\lambda}$
$E_2$ should be given to remove electron (i.e) +ve
$\lambda=\large\frac{6.625\times 10^{-27}\times 3.0\times 10^{10}}{5.425\times 10^{-12}}$
$\Rightarrow 3663.6\times 10^{-8}cm$
$\Rightarrow 3663.6A^{\large\circ}$
Hence (c) is the correct answer.
answered Jan 16, 2014 by sreemathi.v
 

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