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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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The wavelength of $\;K_{\alpha}\;$ X-ray produced by X-ray to be is 0.76$A^{0}\;$. The atomic no.of anticathode material is

$(a)\;82\qquad(b)\;41\qquad(c)\;20\qquad(d)\;10$

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Answer : (b) 41
Explanation :
For $\;K_{\alpha}\;$ X-ray line,
$\large\frac{1}{\lambda_{\alpha}}=R\;(z-1)^2 \;[1-\large\frac{1}{2^2}]$
$=R\;(z-1)^2\;[1-\large\frac{1}{4}]$
$\large\frac{1}{\lambda_{\alpha}}=\large\frac{3}{4}\;R\;(z-1)^2----(1)$
$\lambda_{\alpha}=0.76A^{0}=0.76\times10^{-10}m$
$R=1.097\times10^{7} m$
$(z-1)^2=\large\frac{4}{3}\times\large\frac{1}{0.76\times10^{-10}\times1.09\times10^{7}}$
$\approx 1600$
$z-1=40$
$z=41\;.$
answered Feb 24, 2014 by yamini.v
 

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