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If \lambda_{\mathrm{Cu}} is the wavelength of \mathrm{K}_\alpha X-ray line of copper (atomic number 29) and \lambda_{\mathrm{Mo}} is the wavelength of the \mathrm{K_\alpha}  X-ray line of molybdenum (atomic number 42), then the ratio \mathrm{\lambda_{\mathrm{Cu}} / \lambda_{\mathrm{Mo}}} is close to:
 

Option: 1

1.99

 


Option: 2

2.14


Option: 3

0.50


Option: 4

0.48


Answers (1)

\mathrm{\frac{\lambda_{\mathrm{Cu}}}{\lambda_{\mathrm{Mo}}}=\left(\frac{\mathrm{Z}_{\mathrm{Mo}}-1}{\mathrm{Z}_{\mathrm{Cu}}-1}\right)^2=\frac{41 \times 41}{28 \times 28}=\frac{1681}{784}=2.144 }

Hence option 2 is correct.

Posted by

Sumit Saini

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