(II) An alpha particle (which is a helium nucleus, Q=+2e,\({\bf{m = 6}}{\bf{.64 \times 1}}{{\bf{0}}^{{\bf{ - 27}}}}\;{\bf{kg}}\)) is emitted in a radioactive decay with KE = 5.53 MeV. What is its speed?

The kinetic energy of an object depends on the mass of the object and its speed.

The kinetic energy of an object is given by,

\(KE = \frac{1}{2}m{v^2}\) … (i)

Here, m is the mass and v is the speed.

The kinetic energy of the alpha particle is, \(KE = 5.53\;{\rm{MeV}}\)

Mass of the alpha particle is, \(m = 6.64 \times {10^{ - 27}}\;{\rm{kg}}\)

The kinetic energy of the alpha particle in joule is,

\(\begin{aligned}K{E_1} &= 5.53\;{\rm{keV}}\\ &= \left( {5.53 \times {{10}^6}\;{\rm{eV}}} \right) \times \left( {\frac{{1.6 \times {{10}^{ - 19}}\;{\rm{J}}}}{{1\;{\rm{eV}}}}} \right)\\ &= 8.848 \times {10^{ - 13}}{\rm{J}}\end{aligned}\)

From equation (i), the speed of the alpha particle is calculated as:

\(v = \sqrt {\frac{{2\left( {KE} \right)}}{m}} \)

Substitute the values in the above expression.

\(\begin{aligned}{v_1} &= \sqrt {\frac{{2\left( {8.848 \times {{10}^{ - 13}}\;{\rm{J}}} \right)}}{{6.64 \times {{10}^{ - 27}}\;{\rm{kg}}}}} \\ &= \sqrt {2.6651 \times {{10}^{14}}} \;{\rm{m/s}}\\ &\approx 1.63 \times {10^7}\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of the alpha particle is \(1.63 \times {10^7}\;{\rm{m/s}}\).