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

${\lambda _\alpha } > {{\lambda '}_\alpha } > {\lambda _\beta }$

${{\lambda '}_\alpha } > {\lambda _\beta } > {\lambda _\alpha }$

$\frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{\lambda _\beta }}} = \frac{1}{{{{\lambda '}_\alpha }}}$

NEET UG PHYSICS MCQ: ATOMS

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Analyzing +4 | -1

Let

{\lambda _\alpha },{\lambda _\beta }

and

{{\lambda '}_\alpha }

denote the wavelengths of the

\mathrm{X}

-rays of the

{K_\alpha },{K_\beta }

and

{L_\alpha }

lines in the characteristic

\mathrm{X}

-rays for a metal. Then

View answer and explanation

Correct option: C

\frac{1}{{{\lambda _\beta }}} = \frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{{\lambda '}_\alpha }}}

Option C (

\frac{1}{{{\lambda _\beta }}} = \frac{1}{{{\lambda _\alpha }}} + \frac{1}{{{{\lambda '}_\alpha }}}

) is correct because the energy difference between levels corresponds to the sum of energies of individual transitions.

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