question_answer

Find the momentum and kinetic energy of an electron revolving in the first orbit of hydrogen atom, whose radius is\[0.\text{529}\overset{{}^\circ }{\mathop{\text{A}}}\,\].

Answer:

Momentum \[(P)=?\] Kinetic energy \[(E)=?\] Orbit number \[(n)=1\] Radius of the orbit \[(r)=0.529\overset{0}{\mathop{\text{A}}}\,=0.529\times {{10}^{-10}}m\] \[\frac{Calculation\text{ }of\text{ }momentum\text{ }(mv);}{Terms\text{ }connected\text{ }are\text{ }mv\text{ },\text{ }r\text{ }and\text{ }n}\] \[mvr=\frac{nh}{2\pi }\] \[\Rightarrow \] \[mv=P=\frac{nh}{2\pi }\times \frac{1}{r}\] Substituting the values in the above formula, we get \[\Rightarrow \]\[P=\frac{1\times 6.625\times {{10}^{-34}}}{2\times 3.14}\times \frac{1}{0.529\times {{10}^{-10}}}\] \[={{10}^{-34}}\times \frac{{{10}^{10}}}{0.529}\] \[=\frac{1}{0.529}\times {{10}^{-34}}\times {{10}^{10}}\] \[\approx 1.9\times {{10}^{-24}}kg-m/s\] Therefore the momentum of the electron is \[1.9\times {{10}^{24}}kg-m/s\] We know that kinetic energy \[(E)=\frac{1}{2}m{{v}^{2}}\]                                      ??..(1) Where ?m? is mass of electron \[=9.1\times {{10}^{31}}kg\] How to get velocity (v) in terms of given data? We know that momentum \[P=mv\] \[\Rightarrow \] \[v=\frac{p}{m}\]                                           ............(2) Substituting (2) in (1), we get \[E=\frac{1}{2}\times m\times \frac{{{p}^{2}}}{{{m}^{2}}}\]\[\Rightarrow \]\[E=\frac{{{p}^{2}}}{2m}\]    ............ (3) Substituting the values of ?P? and ?m? in (3), we get \[\Rightarrow \]\[E=\frac{{{p}^{2}}}{2m}\]                                           ............ (3) \[E=\frac{{{(1.9\times {{10}^{-24}})}^{2}}}{2\times 9.1\times {{10}^{-31}}}\] \[=\frac{3.61\times {{10}^{-48}}}{2\times 9.1\times {{10}^{-31}}}\] \[=\frac{3.61\times {{10}^{-48}}\times {{10}^{31}}}{2\times 9.1}\] \[=\frac{3.61}{18.2}\times {{10}^{-17}}\] \[=\frac{3.61}{18.2}\times {{10}^{-18}}=1.98\times {{10}^{-18}}J\] Therefore the energy of the electron is \[1.98\times {{10}^{-18}}J\]