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The wave length λ of Kα-ray line of an anticathode element of atomic number Z is nearly proportional to:

 

  • Option 1)

    Z^{2}

  • Option 2)

    \left ( Z-1 \right )^{2}

  • Option 3)

    \frac{1}{\left ( Z-1 \right )}

  • Option 4)

    \frac{1}{\left ( Z-1 \right )^{2}}

 

Answers (1)

best_answer

 

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-1 \right )\left [ b=1 \right ]or \sqrt{\frac{c}{\lambda }}=a\left ( z-1 \right ) or \frac{c}{\lambda }= a^{2}\left ( z-1 \right )^{2}

\therefore \lambda \alpha \frac{1}{(Z-1)^{2}}

 


Option 1)

Z^{2}

Incorrect option

Option 2)

\left ( Z-1 \right )^{2}

Incorrect option

Option 3)

\frac{1}{\left ( Z-1 \right )}

Incorrect option

Option 4)

\frac{1}{\left ( Z-1 \right )^{2}}

correct option

Posted by

Plabita

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