A) \[{{\lambda }_{\alpha }}>{{{\lambda }'}_{\alpha }}>{{\lambda }_{\beta }}\]
B) \[{{{\lambda }'}_{\alpha }}>{{\lambda }_{\beta }}>{{\lambda }_{\alpha }}\]
C) \[\frac{1}{{{\lambda }_{\beta }}}=\frac{1}{{{\lambda }_{\alpha }}}+\frac{1}{{{{{\lambda }'}}_{\alpha }}}\]
D) \[\frac{1}{{{\lambda }_{\alpha }}}+\frac{1}{{{\lambda }_{\beta }}}=\frac{1}{{{{{\lambda }'}}_{\alpha }}}\]
Correct Answer: C
Solution :
According to the energy diagram of X-ray spectra \[\because \Delta E=\frac{hc}{\lambda }\]\[\Rightarrow \lambda \propto \frac{1}{\Delta E}\] (DE = Energy radiated when e? jumps from, higher energy orbit to lower energy orbit) \[\because {{(\Delta E)}_{{{k}_{\beta }}}}>{{(\Delta E)}_{{{k}_{\alpha }}}}>{{(\Delta E)}_{{{L}_{\alpha }}}}\]\[\therefore {{{\lambda }'}_{\alpha }}>{{\lambda }_{\alpha }}>{{\lambda }_{\beta }}\] Also \[{{(\Delta E)}_{{{k}_{\beta }}}}={{(\Delta E)}_{{{k}_{\alpha }}}}+{{(\Delta E)}_{{{L}_{\alpha }}}}\] \[\Rightarrow \frac{hc}{{{\lambda }_{\beta }}}=\frac{hc}{{{\lambda }_{\alpha }}}+\frac{hc}{{{{{\lambda }'}}_{\alpha }}}\]\[\Rightarrow \frac{1}{{{\lambda }_{\beta }}}=\frac{1}{{{\lambda }_{\alpha }}}+\frac{1}{{{{{\lambda }'}}_{\alpha }}}\]
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