A) greater than \[KE\] of proton
B) zero
C) equal to \[KE\] of proton
D) infinite
Correct Answer: A
Solution :
Key Idea: Relation between kinetic energy \[K\] and momentum\[(p)\]is\[p=\sqrt{2mK}\] The de-Broglie wavelength \[(\lambda )\] is given by \[\lambda =\frac{h}{p}\] ? (i) where, \[h\] is Planck's constant, and \[p\] is momentum. Also\[,\] \[p=\sqrt{2mK}\] ... (ii) From Eqs. (i) and (ii), we get \[\therefore \] \[\lambda =\frac{h}{\sqrt{2mK}}\] \[\Rightarrow \] \[K=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\Rightarrow K\propto \frac{1}{m}\] Hence, kinetic energy of electron will be greater than that of proton because proton has more mass than electron.You need to login to perform this action.
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