1) The energy of an electron having de-Broglie wavelength $\lambda$ is (where, h= planck's constant, m= mass of electron) A) $\frac{h}{2m\lambda}$ B) $\frac{h^{2}}{2m\lambda^{2}}$ C) $\frac{h^{2}}{2m^{2}\lambda^{2}}$ D) $\frac{h^{2}}{2m^{2}\lambda^{}}$ Answer: Option BExplanation:We know that , $\lambda= \frac{h}{\sqrt{2m(KE)}}$ on taking square on both sides,we get $\lambda^{2}= \frac{h^{2}}{2m(KE)}$ $KE = \frac{h^{2}}{2m\lambda^{2}}$
