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If accelerating voltage $V_{0} = \frac{h c R}{e}$ and three times of cut off wave length is equal to the $k_{α}$ wave length of the element then calculate the value of 3z (symbols have their meaning )

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answer is 5.

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Detailed Solution

As electron is accelerated with voltage $V_{0} = \frac{h c R}{e}$ , energy of the electron $e V_{o} = \frac{h c}{λ_{c u t o f f}}$ ,
For $k_{α}$ wavelength of element , $\frac{1}{λ_{0}} = \frac{3 R}{4} {(z - 1)}^{2}$
Given, $3 λ_{c u t o f f} = λ_{α}$
$\Rightarrow 3 \frac{h c}{e V_{0}} = \frac{4}{3 R {(z - 1)}^{2}} \Rightarrow z = 5 / 3 \Rightarrow 3 z = 5$

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If accelerating voltage V0=hcRe and three times of cut off wave length is equal to the kα wave length of the element then calculate the value of 3z (symbols have their meaning )