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A hydrogen atom in the ground state is hit by a particle exciting the electron to 4s orbital.The electron then drops to 2s orbital.What is the frequency of radiation emitted and absorbed in the process?

$\begin{array}{1 1}(a)\;6.15\times 10^{15}Hz&(b)\;7\times 10^{14}Hz\\(c)\;3\times 10^{13}Hz&(d)\;3.5\times 10^{12}Hz\end{array}$

1 Answer

Energy is absorbed when electron moves from ground state (n=1)+0 $4s$ orbital (n=4)
$\Delta E=13.6Z^2\big(\large\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2}\big)$$eV$
Putting $n_1=1$ and $n_2=4$ we get,
$\Delta E=12.75eV=12.75(1.6\times 10^{-19})J=2.04\times 10^{-18}J$
Frequency =$\large\frac{\Delta E}{h}=\frac{2.04\times 10^{-18}}{6.63\times 10^{-34}}$
$\Rightarrow 3.07\times 10^{15}$Hz
When electron drops from n=4 to n=2(2s) energy is emitted and is given by the same relation.
Put $n_1=2$ and $n_2=4$ in the expression of $\Delta E$
$\Delta E=2.55eV=4.08\times 10^{-19}J$
Frequency =$\large\frac{\Delta E}{h}$$=6.15\times 10^{15}Hz$
Hence (a) is the correct option.
answered Jan 9, 2014 by sreemathi.v

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