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

The kinetic energy of a charged particle is given by,

\(KE = \frac{1}{2}m{v^2}\) 鈥� (i)

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

The kinetic energy of the electron in the first case is, \(K{E_1} = 850\;{\rm{eV}}\)

The kinetic energy of the electron in the second case is, \(K{E_1} = 0.50\;{\rm{keV}} = 500\;{\rm{eV}}\)

Mass of the electron is, \(m = 9.11 \times {10^{ - 31}}\;{\rm{kg}}\)

The kinetic energy of the electron in joule is,

\(\begin{aligned}K{E_1} &= 850\;{\rm{eV}}\\ &= \left( {850\;{\rm{eV}}} \right) \times \left( {\frac{{1.6 \times {{10}^{ - 19}}\;{\rm{J}}}}{{1\;{\rm{eV}}}}} \right)\\ &= 1.36\; \times {10^{ - 16}}{\rm{J}}\end{aligned}\)

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

\({v_1} = \sqrt {\frac{{2\left( {K{E_1}} \right)}}{m}} \)

Substitute the values in the above expression.

\(\begin{aligned}{v_1} &= \sqrt {\frac{{2\left( {1.36 \times {{10}^{ - 16}}\;{\rm{J}}} \right)}}{{9.11 \times {{10}^{ - 31}}\;{\rm{kg}}}}} \\ &= \sqrt {2.9857 \times {{10}^{14}}} \;{\rm{m/s}}\\ &= 1.73 \times {10^7}\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of an electron with kinetic energy 850 eV is \(1.73 \times {10^7}\;{\rm{m/s}}\).

The kinetic energy of the electron in joule is,

\(\begin{aligned}K{E_2} &= 500\;{\rm{eV}}\\ &= \left( {500\;{\rm{eV}}} \right) \times \left( {\frac{{1.6 \times {{10}^{ - 19}}\;{\rm{J}}}}{{1\;{\rm{eV}}}}} \right)\\ &= 8.0\; \times {10^{ - 17}}{\rm{J}}\end{aligned}\)

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

\({v_2} = \sqrt {\frac{{2\left( {K{E_2}} \right)}}{m}} \)

Substitute the values in the above expression.

\(\begin{aligned}{v_2} &= \sqrt {\frac{{2\left( {8.0 \times {{10}^{ - 17}}\;{\rm{J}}} \right)}}{{9.11 \times {{10}^{ - 31}}\;{\rm{kg}}}}} \\ &= \sqrt {1.7563 \times {{10}^{14}}} \;{\rm{m/s}}\\ &= 1.32 \times {10^7}\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of an electron with the kinetic energy of 0.50 keVis \(1.32 \times {10^7}\;{\rm{m/s}}\).