If the mass of a neutron is \( 1.7\times10^{-27}\,\mathrm{kg} \), then the de Broglie wavelength of a neutron of energy 3 eV is (take \( h=6.6\times10^{-34}\,\mathrm{J\,s} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.6\times10^{-11}\,\mathrm{m} \)
The de Broglie wavelength is \( \lambda=h/p \). For a particle of mass \( m \) and kinetic energy \( K \), the momentum is \( p=\sqrt{2mK} \), with \( 1\,\mathrm{eV}=1.6\times10^{-19}\,\mathrm{J} \).
\( \begin{aligned} K &= 3\,\mathrm{eV} \\ &= 3\times1.6\times10^{-19}\,\mathrm{J} \\ &= 4.8\times10^{-19}\,\mathrm{J} \end{aligned} \)
\( \begin{aligned} p &= \sqrt{2mK} \\ &= \sqrt{2\times1.7\times10^{-27}\,\mathrm{kg}\times4.8\times10^{-19}\,\mathrm{J}} \\ &\approx 4.04\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} \end{aligned} \)
\( \begin{aligned} \lambda &= \frac{h}{p} \\ &= \frac{6.6\times10^{-34}\,\mathrm{J\,s}}{4.04\times10^{-23}\,\mathrm{kg\,m\,s^{-1}}} \\ &\approx 1.6\times10^{-11}\,\mathrm{m} \end{aligned} \)