Characteristic X-rays of frequency $4.2 \times 10^{18} H z$ are emitted from a metal due to transition from $L -$ to $K - s h e l l$ . Find the atomic number of the metal using Moseley's law. Take Rydberg constant $R = 1.1 \times 10^{7} m^{- 1}$ .

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Solution

We know that,
$\sqrt{V} = a (z - b)$
now, $\frac{1}{λ} = \frac{V}{c} = \frac{4.2 \times 10^{18}}{3 \times 10^{8}}$
$Z = R {(Z - 1)}^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$Z = 1.1 \times 10^{7} {(Z - 1)}^{2} [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$
$Z = 42$

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Characteristic X-rays of frequency 4.2×1018Hz are emitted from a metal due to transition from L− to K−shell. Find the atomic number of the metal using Moseley's law. Take Rydberg constant R=1.1×107m−1.

We know that,√V=a(z−b)now, 1λ=Vc=4.2×10183×108Z=R(Z−1)2[1n21−1n22]Z=1.1×107(Z−1)2[112−122]Z=42

Characteristic X-rays of frequency $4.2 \times 10^{18} H z$ are emitted from a metal due to transition from $L -$ to $K - s h e l l$ . Find the atomic number of the metal using Moseley's law. Take Rydberg constant $R = 1.1 \times 10^{7} m^{- 1}$ .

We know that,
$\sqrt{V} = a (z - b)$
now, $\frac{1}{λ} = \frac{V}{c} = \frac{4.2 \times 10^{18}}{3 \times 10^{8}}$
$Z = R {(Z - 1)}^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
$Z = 1.1 \times 10^{7} {(Z - 1)}^{2} [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$
$Z = 42$