Question

Let ${\lambda }_{\alpha },{\lambda }_{\beta }$ and ${\lambda }_{\alpha }^{\prime }$ denote the wavelengths of the X-rays of the $K_{\alpha },K_{\beta }$ and $L_{\alpha }$ lines 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)

The correct options are
A ${\lambda }_{\alpha }^{\prime }>{\lambda }_{\alpha }>{\lambda }_{\beta }$
C $\frac{1}{{\lambda }_{\beta }}=\frac{1}{{\lambda }_{\alpha }}+\frac{1}{{\lambda }_{\alpha }^{\prime }}$

By definition
$\frac{1}{{\lambda }_{\alpha }}=R(1-\frac{1}{2^2})=0.75R$ ----------(1)
$\frac{1}{{\lambda }_{\beta }}=R(1-\frac{1}{3^2})=0.889R$ -----------(2)
$\frac{1}{{\lambda }_{\alpha }^{\prime }}=R(\frac{1}{2^2}-\frac{1}{3^2})=0.139R$ ---------------(3)
$\frac{1}{{\lambda }_{\alpha }}+\frac{1}{{\lambda }_{\alpha }^{\prime }}=\frac{1}{{\lambda }_{\beta }}$
Also
It can be clearly seen from (1), (2) and (3) that
${\lambda }_{\alpha }^{\prime }>{\lambda }_{\alpha }>{\lambda }_{\beta }$