Question

The wavelength of the $k_a$ line for an element of atomic number $43$ is $\lambda$. Then the wavelength of the $K_a$ 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

The correct option is C $\left(\frac{9}{4}\right)\lambda$

As, $\sqrt{V}=a(Z-b)$
$\sqrt{V}=a(Z-1)$
$V=a^2(Z-1{)}^2$
$\frac{C}{\lambda }=a^2(Z-1{)}^2$
$\lambda =\frac{c}{a^2(Z-1{)}^2}$
$z=43$, wavelength $=\lambda$
$\lambda =\frac{c}{a^2(43-1{)}^2}$
$\lambda =\frac{c}{a^2\times 42}$
for $z=29$, wavelength $={\lambda }^{\prime }$
${\lambda }^{\prime }=\frac{C}{a^2(29-1{)}^2}$
${\lambda }^{\prime }=\frac{c}{28a^2}$
$\frac{{\lambda }^{\prime }}{\lambda }=-{\left(\frac{42}{28}\right)}^2={\left(\frac{3}{2}\right)}^2$
${\lambda }^{\prime }=\left(\frac{9}{4}\right)\lambda$.