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 Plancks constant, and p is momentum. Also, \[p=\sqrt{2\,m\,K}\] ?(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