P=mv\] and \[v=\sqrt{\FRAC{2QV}{m}}\] Þ\[p=\sqrt{2QmV}\]Þ\[p\propto \sqrt{Qm}\]Þ\[\frac{{{p}_{E}}}{{{p}_{\alpha }}}=\sqrt{\frac{e\times {{m}_{e}}}{2e\times {{m}_{\alpha }}}}=\sqrt{\frac{{{m}_{e}}}{2{{m}_{\alpha }}}}\]

"> P=mv\] and \[v=\sqrt{\FRAC{2QV}{m}}\] Þ\[p=\sqrt{2QmV}\]Þ\[p\propto \sqrt{Qm}\]Þ\[\frac{{{p}_{E}}}{{{p}_{\alpha }}}=\sqrt{\frac{e\times {{m}_{e}}}{2e\times {{m}_{\alpha }}}}=\sqrt{\frac{{{m}_{e}}}{2{{m}_{\alpha }}}}\]

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The ratio of momenta of an electron and an \[\alpha -\]particle which are accelerated from rest by a potential difference of 100 V is [MNR 1994; RPET 1997]

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) 1 year ago

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Momentum \[P=mv\] and \[v=\sqrt{\FRAC{2QV}{m}}\] Þ\[p=\sqrt{2QmV}\]Þ\[p\propto \sqrt{Qm}\]Þ\[\frac{{{p}_{E}}}{{{p}_{\alpha }}}=\sqrt{\frac{e\times {{m}_{e}}}{2e\times {{m}_{\alpha }}}}=\sqrt{\frac{{{m}_{e}}}{2{{m}_{\alpha }}}}\]

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Posted on 19 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Physics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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