Question

(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?

Short answer

The speed of thealpha particle is \(1.63 \times {10^7}\;{\rm{m/s}}\).

Step-by-step solution

Understanding of kinetic energy

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.

Given information

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

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

Determination of the speed of the alpha particle

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 \(1.63 \times {10^7}\;{\rm{m/s}}\).