A proton moves with a velocity equal to \( \left(\frac{1}{20}\right)^{\mathrm{th}} \) of velocity of light. The associated de-Broglie wavelength is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2.65 \times 10^{-14}\,\mathrm{m} \)
The de Broglie wavelength of a particle is \( \lambda = h/p = h/mv \). At \( c/20 \) the proton is non-relativistic, so \( p = mv \).
\( \displaystyle v = \frac{3\times10^{8}}{20} = 1.5\times10^{7}\,\mathrm{m\,s^{-1}} \)
\( h = 6.63\times10^{-34}\,\mathrm{J\,s} \)
\( m_p = 1.67\times10^{-27}\,\mathrm{kg} \)
\( \displaystyle \lambda = \frac{6.63\times10^{-34}}{1.67\times10^{-27} \times 1.5\times10^{7}} = \frac{6.63\times10^{-34}}{2.505\times10^{-20}} \)
\( \displaystyle \lambda \approx 2.65\times10^{-14}\,\mathrm{m} \)