Question

The wavelength of the $K_{\alpha }$ line for an element of atomic number $43$ is $\lambda$. Then the wavelength of the $K_{\alpha }$ line for an element of atomic number $29$, is

• A. $\left(\frac{43}{29}\right)\lambda$
• B. $\left(\frac{42}{28}\right)\lambda$
• C. $\left(\frac{9}{4}\right)\lambda$
• D. $\left(\frac{4}{9}\right)\lambda$

Answer 1

Answer: (C)

Moseley's law is given by
$\sqrt{v}=a(Z-b)$
For $K_{\alpha }$ line, $b=1$
$\sqrt{v}=a(Z-1)$
Squaring, we get
$v=a^2(Z-1{)}^2$
or$\frac{c}{\lambda }=a^2(Z-1{)}^2$
or$\lambda =\frac{c}{a^2(Z-1{)}^2}$
For $Z=43$, wavelength $=\lambda$
$\therefore \lambda =\frac{c}{a^2(43-1{)}^2}$
or $\lambda =\frac{c}{a^2(42{)}^2}...(i)$
For $Z=29$, wavelength $={\lambda }^{\prime }$
${\lambda }^{\prime }=\frac{c}{a^2(29-1{)}^2}$
or ${\lambda }^{\prime }=\frac{c}{a^2(28{)}^2}...(ii)$
Dividing Eq. (ii) by Eq. (i), we get
$\frac{{\lambda }^{\prime }}{\lambda }={\left(\frac{42}{28}\right)}^2={\left(\frac{3}{2}\right)}^2$
$\therefore {\lambda }^{\prime }=\left(\frac{9}{4}\right)\lambda$