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

The angular momentum of an electron in a Bohr's orbit of H-atom is $4.2178\times 10^{-34}Kg-m^2/s$.Calculate the spectral line emitted when electron falls from this level to next lower level.

$\begin{array}{1 1}(a)\;1.8\times 10^{-2}cm&(b)\;1.8\times 10^{-4}cm\\(c)\;2.8\times 10^{-4}cm&(d)\;1.7\times 10^{-2}cm\end{array}$

1 Answer

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$\therefore \large\frac{nh}{2\pi}$$=4.2178\times 10^{-34}$
Or $n=\large\frac{4.2178\times 10^{-34}\times 2\times 3.14}{6.625\times 10^{-34}}$
$\Rightarrow 4$
Thus $\large\frac{1}{\lambda}=$$R_H\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
The transition spectral line for $4^{th}$ to $3^{rd}$ shell is
$\Rightarrow \lambda=1.8\times 10^{-4}cm$
Hence (b) is the correct answer.
answered Jan 16, 2014 by sreemathi.v

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