A particle is associated with a de Broglie wave of wavelength 1.2 nm. The kinetic energy of the particle is made four times. Calculate the new value…

\[\begin{aligned} \lambda & =\frac{\mathrm{h}}{\sqrt{2 m K}} \\ \lambda^{\prime} & =\frac{\mathrm{h}}{\sqrt{2 m \times}} \\ & =\frac{1}{2} \\ & =\frac{\lambda}{2} \\ & =\frac{1.2 n m}{2} \\ & =0.6 \mathrm{~nm} \end{aligned} \]