The de Broglie wavelength associated with a proton moving with a speed of 1 m/s will be: [Plank's Constant = $6.63 × 10^{-34}Js, Mp=1.6×10^{-27}kg$]
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $4.14×10^{-7}m$
De-Broglie wavelength (λ) is -
$λ=\frac{h}{P}=\frac{h}{mv}$ [P = Momentum]
$∴λ=\frac{6.626×10^{-34}}{(1.6×10^{-27})×1}=4.14×10^{-7}m$