For what kinetic energy of proton, will the associated de Broglie wavelength be 16.5 nm? (Given $m_p=1.675×10^{-27}kg,\,h=6.63×10^{-34}Js$)

$5.2×10^{-4}J$ $5.2×10^{-20}J$ $4.8×10^{-25}J$ $4.8×10^{-30}J$

$4.8×10^{-25}J$

Kinetic energy of proton, $K=\frac{1}{2}mv^2$ $=\frac{1}{2m}(mv^2)=\frac{1}{2m}(\frac{h^2}{λ^2})$  $(∵λ=\frac{h}{p})$ $=\frac{(6.63×10^{-34})^2}{2×1.675×10^{-27}×(16.5×10^{-9})^2}=4.8×10^{-25}J$