Question

Let ${\lambda }_{\alpha }$, ${\lambda }_{\beta }$ and ${\lambda }_{\alpha }^{{}^{\prime }}$, denote the wavelengths of the X-rays of the $K_{\alpha }$, $K_{\beta }$ and $L_{\alpha }$ line in the characteristic X-rays for a metal. Then

• A. ${\lambda }_{\alpha }^{{}^{\prime }}>{\lambda }_{\alpha }>{\lambda }_{\beta }$
• B. ${\lambda }_{\alpha }^{{}^{\prime }}>{\lambda }_{\beta }>{\lambda }_{\alpha }$
• C. $\frac{1}{{\lambda }_{\beta }}=\frac{1}{{\lambda }_{\alpha }}+\frac{1}{{\lambda }_{\alpha }^{{}^{\prime }}}$
• D. $\frac{1}{{\lambda }_{\alpha }}+\frac{1}{{\lambda }_{\beta }}=\frac{1}{{\lambda }_{\alpha }^{{}^{\prime }}}$

Answer 1

Answer: (A), (C)

${\lambda }_{\alpha }^{{}^{\prime }}>{\lambda }_{\alpha }>{\lambda }_{\beta }$
$\frac{1}{{\lambda }_{\beta }}=\frac{1}{{\lambda }_{\alpha }}+\frac{1}{{\lambda }_{\alpha }^{{}^{\prime }}}$

For $K_{\alpha }$ line : $E_K-E_L=\frac{hc}{{\lambda }_{\alpha }}....\left(1\right)$

For $K_{\beta }$ line : $E_K-E_M=\frac{hc}{{\lambda }_{\beta }}....\left(2\right)$

For $L_{\alpha }$ line : $E_L-E_M=\frac{hc}{{\lambda }_{\alpha }^{\prime }}....\left(3\right)$

As $E_L-E_M<E_K-E_L<E_K-E_M$ so, ${\lambda }_{\alpha }^{\prime }>{\lambda }_{\alpha }>{\lambda }_{\beta }$

From (2), $(E_K-E_L)+(E_L-E_M)=\frac{hc}{{\lambda }_{\beta }}$

using (1) and (2), $\frac{1}{{\lambda }_{\alpha }^{\prime }}+\frac{1}{{\lambda }_{\alpha }}=\frac{1}{{\lambda }_{\beta }}$