Question

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 molybdenum (atomic number 42), then the ratio ${\lambda }_{cu}/{\lambda }_{Mo}$ is close to :

• A. $1.99$
• B. $2.14$
• C. $0.50$
• D. $0.48$

Answer 1

Answer: (B)

$2.14$

We know,

The relation between $\lambda$ and $Z$ is given by,

$\lambda =(Z-1{)}^{-2}$

$\therefore \frac{{\lambda }_{cu}}{{\lambda }_{mo}}=(\frac{Z_{mo}-1}{Z_{cu}-1}{)}^2$

$\therefore \frac{{\lambda }_{cu}}{{\lambda }_{mo}}=(\frac{42-1}{29-1}{)}^2=2.144$