Question

(II) What is the speed of an electron with kinetic energy (a) 850 eV, and (b) 0.50 keV?

Short answer

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

(b) The speed of electron with kinetic energy 0.50 keV is \(1.32 \times {10^7}\;{\rm{m/s}}\).

Step-by-step solution

Understanding of kinetic energy

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,

\(KE = \frac{1}{2}m{v^2}\) … (i)

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

Given information

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, \(m = 9.11 \times {10^{ - 31}}\;{\rm{kg}}\)

(a) Determination of speed of electron with kinetic energy 850 eV

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 \(1.73 \times {10^7}\;{\rm{m/s}}\).

(b) Determination of speed of electron with kinetic energy 0.50 keV

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 \(1.32 \times {10^7}\;{\rm{m/s}}\).