Question

Target of a coolidge tube contains two impurities. Atomic number of the target is Z and that of $1^{st}$ and $2^{nd}$ impurities is $Z_1\text{and}{Z}_2.$ Wavelength of $K_{\alpha }$ lines of target is $\lambda$ and that of impurities are ${\lambda }_1$ and ${\lambda }_2$ where, $\frac{\lambda }{{\lambda }_1}=4$ and $\frac{\lambda }{{\lambda }_2}=\frac{1}{4}$, then

• A. $Z_1=(2Z-1)$
• B. $Z_1=(2Z+1)$
• C. $Z_2=\frac{(Z+1)}{2}$
• D. $Z_2=(\frac{Z}{2}-1)$

Answer 1

Answer: (A), (C)

The correct options are
A $Z_1=(2Z-1)$
C $Z_2=\frac{(Z+1)}{2}$

$\sqrt{\frac{1}{\lambda }}=\sqrt{\frac{3}{4}R}(Z-1)$ ...(1)
$\sqrt{\frac{1}{{\lambda }_1}}=\sqrt{\frac{3R}{4}}(Z_1-1)$ ...(2)
$\sqrt{\frac{1}{{\lambda }_2}}=\sqrt{\frac{3R}{4}}(Z_2-1)$ ...(3)
divide (2) by (1)
$\sqrt{\frac{\lambda }{{\lambda }_1}}=\frac{(Z_1-1)}{Z-1}$
$2Z-2=Z_1-1$
$Z_1=(2Z-1)$
In same way dividing (3) by (1),
we get $Z_2=\frac{Z+1}{2}$