Question

An element with atomic number $Z=11$ emits $K_a$ X-ray of wavelength $\lambda$, then the atomic number of the element which emits $K_a$. X-ray of wavelength $4\lambda$ is?

• A. $11$
• B. $44$
• C. $6$
• D. $5$

Answer 1

Answer: (C)

$6$

According to Moseley's law
$\sqrt{v}=a(Z-b)$
Squaring both sides, $v=a^2(Z-b{)}^2$
or $\frac{c}{\lambda }=a^2(Z-b{)}^2$
Therefore, for two different elements, the ratio of wavelengths is given by
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{(Z_2-1{)}^2}{(Z_1-1{)}^2}$
Given : ${\lambda }_1=\lambda ,Z_1=11,{\lambda }_2=4\lambda ,Z_2=?$
$\therefore \frac{\lambda }{4\lambda }=\frac{(Z_2-1{)}^2}{(11-1{)}^2};$
or $(Z_2-1{)}^2=25;$
$\therefore Z_2=6$.