Discussion Forum

If $\lambda_{Cu}$  is the wavelength of  $k_{\alpha}$ , X-ray line of copper (atomic number 29)  and $\lambda_{Mo}$  is the wavelength of the   $k_{\alpha}$ , X-ray line of molybdnenum (atomic number 42), then  the ratio  $\lambda_{Cu}$/ $\lambda_{Mo}$  is close to 

Answer:

Explanation:

$k_{\alpha}$   transition takes place from n₁=2 to n₂=1

$\therefore$     $  \frac{1}{\lambda}=R(Z-b)^{2}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$

 For K-series, b=1

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

 $\Rightarrow$        $ \frac{\lambda_{Cu}}{\lambda_{Mo}}=\frac{(z_{M0}-1)^{2}}{(z_{Cu}-1)^{2}}$

                     =$\frac{(42-1)^{2}}{(29-1)^{2}}=\frac{41\times41}{28\times 28}$

                       $=\frac{1681}{784}=2.144$