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.
You will be redirected in
3 sec
