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

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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.

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