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.

Answer:

            (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.