A) 8V
B) 6V
C) 4V
D) 12V
Correct Answer: A
Solution :
de-Broglie gave the following relation between momentum (p) and wavelength \[(\lambda )\] as \[\lambda =\frac{h}{p}\] where h is Plancks constant. Also, relation between momentum (p) and kinetic energy \[(qV)\] is \[p=\sqrt{2\,m\,q\,V}\] where q is charge and m. is mass. we have \[{{\lambda }_{p}}={{\lambda }_{\alpha }}\] (given) \[\frac{h}{\sqrt{2\,{{m}_{p}}\,{{q}_{p}}\,{{V}_{p}}}}=\frac{h}{\sqrt{2\,{{m}_{\alpha }}\,{{q}_{\alpha }}\,\,\,{{V}_{\alpha }}}}\] \[\Rightarrow \] \[{{m}_{p}}\,{{q}_{p}}\,{{V}_{p}}={{m}_{\alpha }}\,{{q}_{\alpha }}\,\,{{V}_{a}}\] \[({{V}_{\alpha }}=V)\] \[\Rightarrow \] \[{{V}_{p}}=\left( \frac{{{m}_{\alpha }}}{{{m}_{p}}} \right)\,\,\left( \frac{{{q}_{\alpha }}}{{{q}_{p}}} \right)\,\,{{V}_{\alpha }}\] \[\therefore \] \[{{V}_{p}}=4\times 2\,{{V}_{\alpha }}\] \[\Rightarrow \] \[{{V}_{p}}=8\] volt
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