question_answer

Suppose that a material emits X-rays of wavelengths\[\,\lambda {{\kappa }_{\alpha }},\lambda {{\kappa }_{\beta }},{{\lambda }_{{{L}_{\alpha }}}}\], when it is excited by fast moving electrons; the wavelengths corresponding to \[{{K}_{\alpha }},{{K}_{\beta }},{{L}_{\alpha }}\]X- rays of the material respectively. Then we can write

A)  \[\,\lambda {{\kappa }_{\beta }}=\lambda {{\kappa }_{\alpha }}+{{\lambda }_{{{L}_{\alpha }}}}\]

B)  \[\sqrt{\,\lambda {{\kappa }_{\beta }}}=\sqrt{\lambda {{\kappa }_{\alpha }}}+\sqrt{{{\lambda }_{{{L}_{\alpha }}}}}\]

C)  \[\,\frac{1}{\lambda {{\kappa }_{\beta }}}=\frac{1}{\lambda {{\kappa }_{\alpha }}}+\frac{1}{{{\lambda }_{{{L}_{\alpha }}}}}\]

D)  \[\,\frac{1}{\sqrt{\lambda {{\kappa }_{\beta }}}}=\frac{1}{\sqrt{\lambda {{\kappa }_{\alpha }}}}+\frac{1}{\sqrt{{{\lambda }_{{{L}_{\alpha }}}}}}\]

Correct Answer: C

Solution :

[c] The energy level diagram of the atom is shown in the figure. It is clear that \[{{E}_{{{\kappa }_{\beta }}}}={{E}_{\kappa \alpha }}+{{E}_{{{L}_{\alpha }}}}\] \[or,\text{ }{{v}_{{{\kappa }_{\beta }}}}={{v}_{\kappa \alpha }}+{{v}_{{{L}_{\alpha }}}}\] \[or,\frac{1}{{{\lambda }_{{{\kappa }_{\beta }}}}}=\frac{1}{{{\lambda }_{\kappa \alpha }}}+\frac{1}{{{\lambda }_{{{L}_{\alpha }}}}}\]