A) \[\frac{{{\lambda }_{e}}}{{{\lambda }_{\operatorname{P}}}}=\sqrt{\frac{{{\operatorname{m}}_{\operatorname{p}}}}{{{\operatorname{m}}_{\operatorname{e}}}}}\]
B) \[\frac{{{\lambda }_{e}}}{{{\lambda }_{\operatorname{P}}}}=\sqrt{\frac{{{\operatorname{m}}_{e}}}{{{\operatorname{m}}_{\operatorname{P}}}}}\]
C) \[\frac{{{\lambda }_{e}}}{{{\lambda }_{\operatorname{P}}}}=\frac{1}{2}\sqrt{\frac{{{\operatorname{m}}_{e}}}{{{\operatorname{m}}_{\operatorname{P}}}}}\]
D) \[\frac{{{\lambda }_{e}}}{{{\lambda }_{\operatorname{P}}}}=2\sqrt{\frac{{{\operatorname{m}}_{\operatorname{P}}}}{{{\operatorname{m}}_{e}}}}\]
Correct Answer: D
Solution :
Energy in joule (E) = charge \[\times \]potential diff. in volt \[{{E}_{electron}}={{q}_{e}}V\]and\[{{E}_{proton}}={{q}_{p}}4V\] de-Broglie wavelength\[\lambda =\frac{h}{P}=\frac{h}{\sqrt{2mE}}\] \[{{\lambda }_{e}}=\frac{h}{\sqrt{2{{m}_{e}}eV}}\]and\[{{\lambda }_{p}}=\frac{h}{\sqrt{2{{m}_{P}}e4V}}\] \[(\because {{q}_{e}}={{q}_{P}})\] \[\therefore \]\[\frac{{{\lambda }_{e}}}{{{\lambda }_{P}}}=\frac{\frac{h}{\sqrt{2{{m}_{e}}eV}}}{\frac{h}{\sqrt{2{{m}_{P}}e4V}}}=\sqrt{\frac{2{{m}_{P}}e4V}{2{{m}_{e}}eV}}\] \[=2\sqrt{\frac{{{m}_{P}}}{{{m}_{e}}}}\] \[\therefore \] Greatest amount of heat generated by S.
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