(a) The mass and wavelength of a moving electron are 9.0 x 10\(^{-31}\) kg and 1.0 x 10\(^{-10}\)m respectively. Calculate the kinetic energy of the electro...
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(a) The mass and wavelength of a moving electron are 9.0 x 10\(^{-31}\) kg and 1.0 x 10\(^{-10}\)m respectively. Calculate the kinetic energy of the electron. [ h = 6.6 x 10\(^{-34}\) Js]
Answer Details
Kinetic energy of the electron from its de Broglie wavelength
The de Broglie relation links wavelength to momentum:
\[ \lambda = \frac{h}{mv} \quad\Rightarrow\quad v = \frac{h}{m\lambda} \]Substituting \(h = 6.6\times10^{-34}\,\text{Js}\), \(m = 9.0\times10^{-31}\,\text{kg}\), \(\lambda = 1.0\times10^{-10}\,\text{m}\):
\[ v = \frac{6.6\times10^{-34}}{(9.0\times10^{-31})(1.0\times10^{-10})} = \frac{6.6\times10^{-34}}{9.0\times10^{-41}} = 7.33\times10^{6}\,\text{ms}^{-1} \]Then the kinetic energy is:
\[ E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(9.0\times10^{-31})(7.33\times10^{6})^2 \] \[ E_k = \tfrac{1}{2}(9.0\times10^{-31})(5.38\times10^{13}) \approx 2.42\times10^{-17}\,\text{J} \]Equivalently, using \(E_k = \dfrac{h^2}{2m\lambda^2}\) gives the same result, \(E_k \approx 2.4\times10^{-17}\,\text{J}\).
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