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If the K_{\alpha }  radiation of M0 (z=42) has a wavelength of 0.71A0. Find the wavelength of corresponding radiation of Cu (z=29)

Option: 1

1A^{\circ}


Option: 2

2A^{\circ}


Option: 3

1.52A^{\circ}


Option: 4

1.25A^{\circ}


Answers (1)

best_answer

As we learn

Moseley's law -

\sqrt{\nu }=a(z-b)

- wherein

a=\sqrt{\frac{3RC}{4} }

b=1 for

K_{\alpha} \, \, lines

 

 \sqrt{\nu }=a\left ( z-b \right )

or        \sqrt{\frac{c}{\lambda }}=a\left ( z-b \right )

b\simeq 1           

  \Rightarrow \sqrt{\frac{c}{\lambda_{0}}}=a\left ( 42-1 \right )

 \Rightarrow \sqrt{\frac{c}{\lambda_{cu}}}=a\left ( 29-1 \right )

\frac{\lambda _{Cu}}{\lambda _{Mo}}=\left (\frac{41}{28} \right )^{2}\Rightarrow \lambda _{Cu}=\left (\frac{41}{28} \right )^{2}\times 0.71A^{\circ}

\lambda _{Cu}=1.52A^{\circ}

 

 

Posted by

Pankaj

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