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Wavelength of the $K_{\alpha}$ characteristic X-rays of iron and potassium are $1.93\times 10^{-8}$ are $3.737\times 10^{-8}cm$ respectively.What is the atomic number and name of the element for which characteristic $K_{\alpha}$ wavelength is $2.289\times 10^{-8}cm$.

$\begin{array}{1 1}(a)\;22&(b)\;21\\(c)\;24&(d)\;23\end{array}$

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1 Answer

The frequency of emitted $X$-rays is given by
$V\propto z^2$
$\lambda\propto \large\frac{1}{z^2}$
For $Fe$ (at $N_0=26)$
$\Rightarrow \lambda_1\propto \large\frac{1}{(26)^2}$-------(1)
For $K$ (at $N_0=19)$
$\Rightarrow \lambda_2\propto \large\frac{1}{(19)^2}$-------(2)
For $X$ (at $N_0=z)$
$\Rightarrow \lambda_3\propto \large\frac{1}{(z)^2}$-------(3)
By equ (1) and (3)
$\large\frac{\lambda_1}{\lambda_3}=\frac{(z)}{(26)^2}$
$z^2=\large\frac{\lambda_1}{\lambda_3}$$\times (26)^2$
$\Rightarrow \large\frac{1.931\times 10^{-8}}{2.289\times 10^{-8}}$$\times (26)^2$
$z=23.88$
By equ (2) and (3)
$\large\frac{\lambda_2}{\lambda_3}=\frac{z^2}{(19)^2}$
$z^2=\large\frac{\lambda_2}{\lambda_3}$$\times (19)^2$
$\Rightarrow \large\frac{3.737\times 10^{-8}}{2.289\times 10^{-8}}$$(19)^2$
$z=24.28$
$\therefore$ Atomic no of element is 24 chromium.
Hence (c) is the correct answer.
answered Jan 17, 2014 by sreemathi.v
 

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