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 \[{{E}_{{{\kappa }_{\beta }}}}={{E}_{\kappa \alpha }}+{{E}_{{{L}_{\alpha }}}}\] \[or,\text{ }{{v}_{{{\kappa }_{\beta }}}}={{v}_{\kappa \alpha }}+{{v}_{{{L}_{\alpha }}}}\] \[or,\frac{1}{{{\lambda }_{{{\kappa }_{\beta }}}}}=\frac{1}{{{\lambda }_{\kappa \alpha }}}+\frac{1}{{{\lambda }_{{{L}_{\alpha }}}}}\]You need to login to perform this action.
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