Q.4 (c) Calculate velocity and de-Broglie wavelength of alpha particle of energy 1[unit prefix partly illegible in the source scan; read as k]eV. Given: mass of alpha particle as 6.68x10-27 kg.
Best answer
Assumption: The energy unit is partly obscured in the scan. We read it as \(E=1\) keV. The results for 1 MeV are given at the end.
Given: \(E=1\) keV \(=1\times10^3\times1.6\times10^{-19}=1.6\times10^{-16}\) J, \(m=6.68\times10^{-27}\) kg, \(h=6.63\times10^{-34}\) J s.
Velocity: \(E=\tfrac12mv^2\), so
\[ v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2\times1.6\times10^{-16}}{6.68\times10^{-27}}}=\sqrt{4.79\times10^{10}}=2.19\times10^5\ \text{m/s} \]
de Broglie wavelength:
\[ \lambda=\frac{h}{\sqrt{2mE}}=\frac{6.63\times10^{-34}}{\sqrt{2\times6.68\times10^{-27}\times1.6\times10^{-16}}}=\frac{6.63\times10^{-34}}{1.462\times10^{-21}}=4.53\times10^{-13}\ \text{m} \]
\(v\approx2.19\times10^5\) m/s; \(\lambda\approx4.53\times10^{-13}\) m (≈ 0.0045 Å)
Check: \(\lambda=h/(mv)=6.63\times10^{-34}/(6.68\times10^{-27}\times2.189\times10^5)=4.53\times10^{-13}\) m. This agrees.
(If the energy is 1 MeV: \(v\approx6.92\times10^6\) m/s and \(\lambda\approx1.43\times10^{-14}\) m. The non-relativistic treatment is still valid because \(v\ll c\).)
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