Chemistry - Atomic Structure Question with Solution | TestHub

de - Broglie wavelength is given by : $$\lambda =\frac{h}{mv}$$ $$\begin{array}{rl}& \text{ K.E. }=\frac{1}{2}m{v}^2\\ & v^2=\frac{2KE}{m}\\ & v=\sqrt{\frac{2KE}{m}}\end{array}$$ Substituting this in equation (i) $$\lambda =\frac{h}{m}\sqrt{\frac{m}{2KE}}$$ $\lambda =h\sqrt{\frac{1}{2m(\text{ K.E. })}}$ i.e. $\lambda \propto \frac{1}{\sqrt{KE}}$ $\therefore$ when $KE$ become 4 times wavelength become $1/2$.