Answer –
- Let the orbital speed of the electron in the ground state level of an hydrogen atom, n1= 1, be given by v1 . For charge (e) of an electron, v1 is given by the relation –
![\[{{v}_{1}}=\frac{{{e}^{2}}}{{{n}_{1}}4\pi {{\in }_{0}}\left( \frac{h}{2\pi } \right)}=\frac{{{e}^{2}}}{2{{\in }_{0}}h}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9c66f213faf7f00e294d4ce7cb272fec_l3.png "Rendered by QuickLaTeX.com")
Where,
the electronic charge is given by, e = 1.6 × 10−19 C
![\[{{\in }_{0}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3675c3e65e309a8c1ae18df2cf3cf817_l3.png "Rendered by QuickLaTeX.com")
h is the Planck’s constant, given = 6.62 × 10−34 J s
Therefore the orbital speed becomes,
![\[{{v}_{1}}=\frac{{{\left( 1.6\text{ }\times \text{ }{{10}^{-19}} \right)}^{2}}}{2\times ~8.85\text{ }\times \text{ }{{10}^{-12}}\times 6.62\text{ }\times \text{ }{{10}^{-34}}}=0.0218\times {{10}^{8}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8151caab0f3d9e8332a990c4624b56c1_l3.png "Rendered by QuickLaTeX.com")
![\[{{v}_{1}}=2.18\text{ }\times \text{ }{{10}^{6}}m/s\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e52643d390aab46ac72f8f6eea6d9c6e_l3.png "Rendered by QuickLaTeX.com")
Similarly, we can write the relation for the corresponding orbital speed for level n2= 2, given by –
![\[{{v}_{2}}=\frac{{{e}^{2}}}{{{n}_{2}}2{{\in }_{0}}h}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a0aff843c8f776e04d656d4bd9b60b1a_l3.png "Rendered by QuickLaTeX.com")
![\[{{v}_{2}}=\frac{{{\left( 1.6\text{ }\times \text{ }{{10}^{-19}} \right)}^{2}}}{2\times 2\times ~8.85\text{ }\times \text{ }{{10}^{-12}}\times 6.62\text{ }\times \text{ }{{10}^{-34}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ee0f9e0d556a32a77292cff08533a2d5_l3.png "Rendered by QuickLaTeX.com")
![\[{{v}_{2}}=1.09\times {{10}^{6}}m/s\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f9eba591309decf78bd75e9005fc7d99_l3.png "Rendered by QuickLaTeX.com")
Similary the orbital velocity for level n3= 3 is given by –
![\[{{v}_{3}}=\frac{{{e}^{2}}}{{{n}_{3}}2{{\in }_{0}}h}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cd3f03224d3930609134d3f693eb04a3_l3.png "Rendered by QuickLaTeX.com")
![\[{{v}_{3}}=\frac{{{\left( 1.6\text{ }\times \text{ }{{10}^{-19}} \right)}^{2}}}{3\times 2\times ~8.85\text{ }\times \text{ }{{10}^{-12}}\times 6.62\text{ }\times \text{ }{{10}^{-34}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-361460868ee926cea8acfb330971a4af_l3.png "Rendered by QuickLaTeX.com")
Therefore, the speed of the electron for n=1, n=2, and n=3 is 2.18×106m/s, 1.09×106m/s and 7.27 × 105 m/s in a hydrogen atom.
b)
Suppose, T1 represents the orbital period of the electron when it is in the energy level n1= 1.
We know that the Orbital period is related to orbital speed as:
![\[{{T}_{1}}=\frac{2\pi {{r}_{1}}}{{{v}_{1}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-54d7f0f85653d95589645aa209602812_l3.png "Rendered by QuickLaTeX.com")
Where r1 represents the radius of the orbit given by
![\[{{r}_{1}}=\frac{{{n}_{1}}^{2}{{h}^{2}}{{\in }_{0}}}{\pi m{{e}^{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-acb39013f1c04fb97f188261a38f7610_l3.png "Rendered by QuickLaTeX.com")
Where,
h is the Planck’s constant = 6.62 × 10−34 J s
e is the Charge on an electron = 1.6 × 10−19 C
ε0 is the Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
m is the Mass of an electron = 9.1 × 10−31 kg
So, orbital period is given by –
![\[{{T}_{1}}=\frac{2\pi }{{{v}_{1}}}\times \frac{{{n}_{1}}^{2}{{h}^{2}}{{\in }_{0}}}{\pi m{{e}^{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fc74338ae451eea637c806ca988938dd_l3.png "Rendered by QuickLaTeX.com")
![\[{{T}_{1}}=\frac{2\pi \times {{\left( 1 \right)}^{2}}\times {{\left( 6.62\text{ }\times \text{ }{{10}^{-34}} \right)}^{2}}\times 8.85\text{ }\times \text{ }{{10}^{-12}}}{2.18\times {{10}^{6}}\times \pi \times 9.1\times {{10}^{-31}}\times {{\left( 1.6\text{ }\times \text{ }{{10}^{-19}}~ \right)}^{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0f14e5eb659c87692c99658637be09c3_l3.png "Rendered by QuickLaTeX.com")
![\[{{T}_{1}}=1.527\text{ }\times \text{ }{{10}^{-16}}s\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d970c76f04591147ee906a6b2ad10bc4_l3.png "Rendered by QuickLaTeX.com")
Similarly for level n2 = 2, we can write time period as –
![\[{{T}_{2}}=\frac{2\pi {{r}_{2}}}{{{v}_{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d66409666f57c8e2bd59b41f10f8b621_l3.png "Rendered by QuickLaTeX.com")
Where r2 represents the radius of the orbit given by
![\[{{r}_{2}}=\frac{{{n}_{2}}^{2}{{h}^{2}}{{\in }_{0}}}{\pi m{{e}^{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5a7d019a996e715d87b262839c4aaeb2_l3.png "Rendered by QuickLaTeX.com")
So, we get –
![\[{{T}_{2}}=\frac{2\pi \times {{\left( 2 \right)}^{2}}\times {{\left( 6.62\text{ }\times \text{ }{{10}^{-34}} \right)}^{2}}\times 8.85\text{ }\times \text{ }{{10}^{-12}}}{1.09\times {{10}^{6}}\times \pi \times 9.1\times {{10}^{-31}}\times {{\left( 1.6\text{ }\times \text{ }{{10}^{-19}}~ \right)}^{2}}}=1.22\times {{10}^{-15}}s\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0f5529abe86c05c3e7e5312082c77fed_l3.png "Rendered by QuickLaTeX.com")
Again, similarly proceeding we get the time period for level n3 = 3 –
![\[{{T}_{2}}=\frac{2\pi \times {{\left( 3 \right)}^{2}}\times {{\left( 6.62\text{ }\times \text{ }{{10}^{-34}} \right)}^{2}}\times 8.85\text{ }\times \text{ }{{10}^{-12}}}{7.27\times {{10}^{5}}\times \pi \times 9.1\times {{10}^{-31}}\times {{\left( 1.6\text{ }\times \text{ }{{10}^{-19}}~ \right)}^{2}}}=4.12\times {{10}^{-15}}s\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f5947fbdff84ba54da41f831df3bd440_l3.png "Rendered by QuickLaTeX.com")
Therefore, 1.52×10-16s, 1.22×10-15s and 4.12 × 10-15s are the orbital periods in given levels.