A) V/8 volts
B) V/4 volts
C) 4V volts
D) 8V volts
Correct Answer: A
Solution :
The wavelength associated with a particle of charge q, mass m and accelerated through a potential difference V is given by |
\[\lambda =\frac{h}{\sqrt{2mqV}}\,\,or\,\,V=\frac{{{h}^{2}}}{2mq{{\lambda }^{2}}}\] for proton : \[V=\frac{{{h}^{2}}}{2{{m}_{p}}}{{q}_{p}}{{\lambda }^{2}}\] for \[\alpha \]- particle \[V'=\frac{{{h}^{2}}}{2{{m}_{\alpha }}{{q}_{\alpha }}{{\lambda }^{2}}}\] |
\[(\because {{m}_{\alpha }}=4{{m}_{p}}\,\,and\,\,{{q}_{\alpha }}=2{{q}_{p}})\] |
Thus\[V'=V/8\]. Hence the correct choice is [A]. |
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