Thank you for reporting, we will resolve it shortly
Q.
If $E_1, E_2,$ and $E_3$ represent respectively the kinetic energies of an electron and an alpha particle and a proton each having same de-broglie wavelength then
Structure of Atom
Solution:
Since, $K.E. = \frac{1}{2} mv^{2}$ and $ \lambda = \frac{h}{mv} $
$ \therefore K.E. = \frac{1}{2} m \frac{h^{2}}{m^{2} \lambda^{2}} = \frac{h^{2}}{2m \lambda^{2}} $ As $\lambda $ is the same.
$\therefore K.E. \propto \frac{1}{m} $