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The ionization enthalpy of hydrogen atom is $1.312\times 10^6jmol^{-1}$.The energy required to excite the electron in the atom from $n=1$ to $n=2$ is

$\begin{array}{1 1}(a)\;8.51\times 10^5Jmol^{-1}\\(b)\;6.56\times 10^5Jmol^{-1}\\(c)\;7.56\times 10^5Jmol^{-1}\\(d)\;9.84\times 10^5Jmol^{-1}\end{array}$

1 Answer

$\Delta E=E_2-E_1=-\large\frac{-E_1}{2^2}+\frac{E_1}{1^2}$
$\Rightarrow \large\frac{-1.312\times 10^6}{2^2}$$+1.312\times 10^6$
$\Rightarrow 9.84\times 10^5Jmol^{-1}$
Hence (d) is the correct answer.
answered Jan 13, 2014 by sreemathi.v

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