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 $=\lambda'$
$\lambda'=\frac{c}{a^{2}(29-1)^{2}}$
or $\lambda'=\frac{c}{a^{2}(28)^{2}}\,\,\,...(ii)$
Dividing Eq. (ii) by Eq. (i), we get
$ \frac{\lambda'}{\lambda}=\left(\frac{42}{28}\right)^{2}=\left(\frac{3}{2}\right)^{2} $
$\therefore \lambda'=\left(\frac{9}{4}\right) \lambda$