The frequency of the characterstic X ray of Kα line of metal target 'M' is 2500 cm–1 and the graph between v Vs 'z' is as follows, then atomic number of M is

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The frequency of the characterstic X ray of $K_{α}$ line of metal target 'M' is 2500 cm^–1 and the graph between $\sqrt{v}$ Vs 'z' is as follows, then atomic number of M is
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Detailed Solution

according to the Moseley’s Law, we can write:
$\sqrt{ν} = a (Z - b) \sqrt{ν} = a Z - a b$

where, ab=Intercept=OA=1 ; a=slope=tan45^o=1 ; $ν = 2500 {cm}^{- 1}$

Now, $\sqrt{ν} = \sqrt{2500} = 50$

$50 = 1 (Z) - 1 Z = 51$

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