• # question_answer (i) Using the Bohr's model, calculate the speed of the electron in a H-atom in the n = 1, 2 and 3 levels. (ii) Calculate the orbital period in each of these levels.

(i) Let ${{v}_{1}}$ be the orbital speed of the electron in a H-atom in the ground state level, ${{n}_{1}}=1.$ For charge (e) of an electron, ${{v}_{1}}$ is given by the relation             ${{v}_{1}}=\frac{{{e}^{2}}}{{{n}_{1}}4\pi {{\varepsilon }_{0}}\left( \frac{h}{2\pi } \right)}=\frac{{{e}^{2}}}{2{{\varepsilon }_{0}}h}$ where,   $e=1.6\times {{10}^{-19}}C$ ${{\varepsilon }_{0}}$= permittivity of free space $=8.85\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$ h = Planck's constant$=6.63\times {{10}^{-34}}J-s$ $\therefore$      ${{v}_{1}}=\frac{{{(1.6\times {{10}^{-19}})}^{2}}}{2\times 8.85\times {{10}^{-12}}\times 6.63\times {{10}^{-34}}}$             $=0.0218\times {{10}^{8}}=2.18\times {{10}^{6}}m/s$ We know that,    ${{v}_{n}}={{v}_{1}}/n$ For level ${{n}_{2}}=2,$ we can write the relation for the corresponding orbital speed as,             ${{v}_{2}}=\frac{{{v}_{1}}}{2}=\frac{2.18\times {{10}^{6}}}{2}=1.09\times {{10}^{6}}m/s$ and for level ${{n}_{3}}=3,$ we can write the relation for the  corresponding orbital speed as,             ${{v}_{3}}=\frac{{{v}_{1}}}{3}=\frac{2.18\times {{10}^{6}}}{3}=7.27\times {{10}^{5}}m/s$ Hence, the speed of the electron in a H-atom in n = 1, n = 2 and n = 3 is $2.18\times {{10}^{6}}m/s,$$1.09\times {{10}^{6}}m/s$ and $7.27\times {{10}^{5}}m/s,$respectively, (ii) Let ${{T}_{1}}$ be the orbital period of the electron and given by, $T=2\pi r/v$ where,   r = radius of the orbit $=\frac{{{n}^{2}}{{h}^{2}}{{\varepsilon }_{0}}}{\pi m{{e}^{2}}}$ h = Planck's constant $=6.63\times {{10}^{-34}}J-s$ e = charge on an electron $=1.6\times {{10}^{-19}}C$ ${{\varepsilon }_{0}}$ = permittivity of free space $=8.85\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$ and       m = mass of an electron $=9.1\times {{10}^{-31}}kg$ For        n = 1 $\therefore$      ${{T}_{1}}=\frac{2\pi {{r}_{1}}}{{{v}_{1}}}=\frac{2\times 3.14\times 0.53\times {{10}^{-10}}}{2.182\times {{10}^{6}}}$ As,        ${{T}_{n}}={{n}^{3}}{{T}_{1}}$ $\Rightarrow$   ${{T}_{2}}={{(2)}^{3}}{{T}_{1}}=8\times 1.52\times {{10}^{-16}}$             $=1.22\times {{10}^{-15}}s$ ${{T}_{3}}={{(3)}^{3}}{{T}_{1}}=27\times 1.52\times {{10}^{-16}}=4.10\times {{10}^{-15}}s$ Hence, the orbital period in each of these levels is $1.52\times {{10}^{-16}}s,$$1.22\times {{10}^{-15}}s$ and $4.12\times {{10}^{-15}}s,$ respectively.