For what kinetic energy of a proton will the associated de Broglie wavelength be 16.5 nm? (Given \( m_p=1.675\times10^{-27}\,\mathrm{kg} \), \( h=6.63\times10^{-34}\,\mathrm{J\,s} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4.8\times10^{-25}\,\mathrm{J} \)
The de Broglie wavelength of a particle is \( \lambda=h/p \). A particle of mass \( m \) and kinetic energy \( K \) has momentum \( p=\sqrt{2mK} \). So
\( \displaystyle \lambda=\frac{h}{\sqrt{2mK}} \)
With \( \lambda=16.5\,\mathrm{nm}=16.5\times10^{-9}\,\mathrm{m} \), the kinetic energy is:
\( \begin{aligned} K &= \frac{h^2}{2m\lambda^2} \\ &= \frac{(6.63\times10^{-34}\,\mathrm{J\,s})^2}{2\times1.675\times10^{-27}\,\mathrm{kg}\times(16.5\times10^{-9}\,\mathrm{m})^2} \\ &= \frac{4.40\times10^{-67}}{9.12\times10^{-43}}\,\mathrm{J} \\ &\approx 4.8\times10^{-25}\,\mathrm{J} \end{aligned} \)