question_answer

Electrons are accelerated through a potential difference V and protons are accelerated through a potential difference 4 V. The de-Broglie wavelengths are  \[{{\lambda }_{e}}\] and \[{{\lambda }_{\operatorname{p}}}\] for electrons and protons respectively. The ratio of \[\frac{{{\lambda }_{e}}}{{{\lambda }_{\operatorname{P}}}}\] is given by:              (given\[{{\operatorname{m}}_{e}}\]is mass of electron and\[{{\operatorname{m}}_{\operatorname{p}}}\]is mass of proton)                              [JEE ONLINE 23-04-2013]

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 :

[d] Energy in joule (E) = charge potential diff. in volt and de-Broglie wavelength

and

Greatest amount of heat generated by S.