A) infinite
B) equal to KE of proton
C) zero
D) greater than KE of proton
Correct Answer: D
Solution :
de-Broglie wavelength of a charged particle is given by \[\lambda =\frac{h}{mv}\] ? (i) where \[h\] is Plancks constant. If kinetic energy of particle of mass \[m\] is\[v,\] then \[K=\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow \] \[v=\sqrt{\frac{2K}{m}}\] ?(ii) Combining Eqs. (i) and (ii), we get \[\lambda =\frac{h}{m\sqrt{\frac{2K}{m}}}=\frac{h}{\sqrt{2mK}}\] ?(iii) \[\therefore \] \[{{\lambda }_{e}}=\frac{h}{\sqrt{2{{m}_{e}}{{K}_{e}}}}\] and \[{{\lambda }_{p}}=\frac{h}{\sqrt{2{{m}_{p}}{{K}_{p}}}}\] but \[{{\lambda }_{e}}={{\lambda }_{p}}\] (given) \[\therefore \] \[\frac{h}{\sqrt{2{{m}_{e}}{{K}_{e}}}}=\frac{h}{\sqrt{2{{m}_{p}}{{K}_{p}}}}\] or \[2{{m}_{e}}{{K}_{e}}=2{{m}_{p}}{{K}_{p}}\] or \[\frac{{{K}_{e}}}{{{K}_{p}}}=\frac{{{m}_{p}}}{{{m}_{e}}}\] Since,\[{{m}_{p}}>{{m}_{e}}\]or \[\frac{{{m}_{p}}}{{{m}_{e}}}>1\] \[\therefore \] \[\frac{{{K}_{e}}}{{{K}_{p}}}>1\] \[\Rightarrow \] \[{{K}_{e}}>{{K}_{p}}\] Note: If an electron is accelerated through a potential difference of V volt, then Eq. (iii) takes the form \[\lambda =\frac{h}{\sqrt{2meV}}\] After putting the numerical values for electron, we get \[\lambda =\sqrt{\frac{150}{V}}\overset{\text{o}}{\mathop{\text{A}}}\,.\]
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