The de Broglie wavelength associated with a proton moving with a speed of 1 m/s will be: [Planck's Constant = \( 6.63\times10^{-34}\,\mathrm{J\,s} \), \( M_p = 1.6\times10^{-27}\,\mathrm{kg} \)]
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4.14\times10^{-7}\,\mathrm{m} \)
The de Broglie wavelength of a moving particle is Planck's constant divided by its momentum, \( \lambda = h/p = h/(mv) \).
\( \displaystyle \lambda = \frac{6.63\times10^{-34}\,\mathrm{J\,s}}{1.6\times10^{-27}\,\mathrm{kg}\times1\,\mathrm{m\,s^{-1}}} = 4.14\times10^{-7}\,\mathrm{m} \)