Moseley's equation is represented as $\sqrt{v} = a (Z - b)$ where a and b are constants If $O A = 1$ , then atomic number of the elements showing frequency of $400 H z$ is?

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Solution

The correct option is C 21
Given: $O A = 1 V = 400 H z a = 1$
$\sqrt{V} = a (z - b), \sqrt{400} = 1 (z - 1) 20 = z - 1 z = 21$
Hence atomic number is $21$ .

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Moseley's equation is represented as √v=a(Z−b) where a and b are constants If OA=1, then atomic number of the elements showing frequency of 400Hz is?

The correct option is C 21Given: OA=1V=400Hza=1√V=a(z−b),√400=1(z−1)20=z−1z=21Hence atomic number is 21.

Moseley's equation is represented as $\sqrt{v} = a (Z - b)$ where a and b are constants If $O A = 1$ , then atomic number of the elements showing frequency of $400 H z$ is?

The correct option is C 21
Given: $O A = 1 V = 400 H z a = 1$
$\sqrt{V} = a (z - b), \sqrt{400} = 1 (z - 1) 20 = z - 1 z = 21$
Hence atomic number is $21$ .