A) \[3\sqrt{2}\]
B) \[2\sqrt{2}\]
C) \[2\sqrt{3}\]
D) \[2\sqrt{5}\]
Correct Answer: B
Solution :
[b]: de Broglie wavelength associated with charged particle is given by, or\[\lambda =\frac{h}{\sqrt{2mqV}},\frac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\sqrt{\frac{2{{m}_{\alpha }}{{q}_{\alpha }}V}{2{{m}_{p}}{{q}_{p}}V}}\]where subscripts, p and \[\alpha \] represent proton and \[\alpha \] particle respectively. As\[\frac{{{m}_{p}}}{{{m}_{\alpha }}}=\frac{1}{4}\]and\[\frac{{{q}_{\alpha }}}{{{q}_{p}}}=\frac{2}{1}\Rightarrow \frac{{{\lambda }_{p}}}{{{\lambda }_{\alpha }}}=\sqrt{4\times 2}=2\sqrt{2}\]
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