Question

X-rays emitted from a copper target and a molybdenum target are found to contain a line of wavelength $22.85$ nm attribute to the $K_{\alpha }$ lines of an impure element. The $K_{\alpha }$ lines of copper (Z $=29$) and molybdenum (Z $=42$) have a wavelength $15.42$ nm and $7.12$ nm respectively. Using Moseley's law, $v^{1/2}=a(Z-b)$, calculate the atomic number of the impure element:

• A. $24$
• B. $28$
• C. $30$
• D. $32$

Answer 1

The correct option is A $24$

The moseley's law is $\sqrt{\nu }=\sqrt{\frac{c}{\lambda }}=a(Z-b)$
For copper, $\sqrt{\frac{3\times {10}^8}{15.42\times {10}^{-9}m}}=a(29-b)$......(1)
For, Mo, $\sqrt{\frac{3\times {10}^8}{7.12\times {10}^{-9}m}}=a(42-b)$......(2)
Subtract equation (1) from equation (2)
$6.58\times {10}^7=42a-29a=13a$
$a=5.06\times {10}^6$
Substitute the value of a in equation (1)
$\sqrt{\frac{3\times {10}^8}{15.42\times {10}^{-9}m}}=5.06\times {10}^6(29-b)$
$b=1.44$
For impurity, $\sqrt{\frac{3\times {10}^8}{22.852\times {10}^{-9}m}}=5.06\times {10}^6\times Z(29-1.44)$
$Z=24$