In the Geiger-Marsden experiment, a beam of α-particles of kinetic energy 5.5 MeV fell on a thin gold foil. The speed of these α-particles was about (Take the mass of an α-particle as \(6.6\times10^{-27}\ \mathrm{kg}\) and \(1\ \mathrm{MeV}=1.6\times10^{-13}\ \mathrm{J}\).)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.6\times10^{7}\ \mathrm{m\,s^{-1}}\)
The kinetic energy of a moving particle is \( K=\frac{1}{2}mv^{2} \). So its speed is \( v=\sqrt{2K/m} \). First change the energy from MeV into joules.
\( K=5.5\ \mathrm{MeV} \)
\( m=6.6\times10^{-27}\ \mathrm{kg} \)
\( \begin{aligned} K &= 5.5\times1.6\times10^{-13}\ \mathrm{J} \\ &= 8.8\times10^{-13}\ \mathrm{J} \end{aligned} \)
\( \begin{aligned} v &= \sqrt{\frac{2K}{m}} \\ &= \sqrt{\frac{2\times8.8\times10^{-13}}{6.6\times10^{-27}}} \\ &= \sqrt{2.7\times10^{14}}\ \mathrm{m\,s^{-1}} \\ &\approx 1.6\times10^{7}\ \mathrm{m\,s^{-1}} \end{aligned} \)
Hence, the α-particles moved at about \( 1.6\times10^{7}\ \mathrm{m\,s^{-1}} \). This is about one-twentieth of the speed of light.