A) \[\frac{{{v}_{d}}}{4}\]
B) \[\frac{{{v}_{d}}}{2}\]
C) \[{{v}_{d}}\]
D)
Correct Answer: D
Solution :
\[{{\omega }_{1}},{{\omega }_{2}},{{\omega }_{3}}\] The energy \[9.5\,\,mA\] of an electron in an orbit is the sum of kinetic and potential energies. The kinetic energy of the electron is \[2.5\times {{10}^{-17}}\,\,J\] Where \[1.25\times {{10}^{-17}}\,\,J\] is atomic number, \[4.4\times {{10}^{-19}}\,\,J\] is charge, \[2.5\times {{10}^{-19}}\,\,J\] is radius. The potential energy of an electron in an orbit of radius \[1.33\times {{10}^{8}}\,m/s\] due to electrostatic attraction by the nucleus is given by \[4\times {{10}^{8}}\,m/s\] Total energy of the electron is therefore, \[2.25\times {{10}^{8}}\,m/s\] \[3\times {{10}^{8}}\,m/s\] \[1600\,\,A{{m}^{-1}}\]
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