Category : JEE Main & Advanced
(1) Radius of orbit : For an electron around a stationary nucleus the electrostatics force of attraction provides the necessary centripetal force
i.e. \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(Ze)e}{{{r}^{2}}}=\frac{m{{v}^{2}}}{r}\] ... (i)
also \[mvr=\frac{nh}{2\pi }\] ...(ii)
From equation (i) and (ii) radius of \[{{n}^{th}}\] orbit
\[{{r}_{n}}=\frac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}kZm{{e}^{2}}}=\frac{{{n}^{2}}{{h}^{2}}{{\varepsilon }_{0}}}{\pi mZ{{e}^{2}}}=0.53\frac{{{n}^{2}}}{Z}{\AA}\]
\[\text{(}\,k=\frac{1}{4\pi {{\varepsilon }_{0}}}\text{)}\]
\[\Rightarrow \]\[{{r}_{n}}\propto \frac{{{n}^{2}}}{Z}\]
(2) Speed of electron : From the above relations, speed of electron in nth orbit can be calculated as
\[{{v}_{n}}=\frac{2\pi kZ{{e}^{2}}}{nh}=\frac{Z{{e}^{2}}}{2{{\varepsilon }_{0}}nh}=\left( \frac{c}{137} \right).\frac{Z}{n}=2.2\times {{10}^{6}}\frac{Z}{n}\,m/sec\]
where (c = speed of light \[3\times {{10}^{8}}\] m/s)
Some other quantities for revolution of electron in \[{{n}^{th}}\]orbit
Quantity | Formula | Dependency on n and Z |
(1) Angular speed | \[{{\omega }_{n}}=\frac{{{v}_{n}}}{{{r}_{n}}}=\frac{\pi m{{z}^{2}}{{e}^{4}}}{2\varepsilon _{0}^{2}{{n}^{3}}{{h}^{3}}}\] | \[{{\omega }_{n}}\propto \frac{{{Z}^{2}}}{{{n}^{3}}}\] |
(2) Frequency | \[{{\nu }_{n}}=\frac{{{\omega }_{n}}}{2\pi }=\frac{m{{z}^{2}}{{e}^{4}}}{4\varepsilon _{0}^{2}{{n}^{3}}{{h}^{3}}}\] | \[{{\nu }_{n}}\propto \frac{{{Z}^{2}}}{{{n}^{3}}}\] |
(3) Time period | \[{{T}_{n}}=\frac{1}{{{\nu }_{n}}}=\frac{4\varepsilon _{0}^{2}{{n}^{3}}{{h}^{3}}}{m{{z}^{2}}{{e}^{4}}}\] | \[{{T}_{n}}\propto \frac{{{n}^{3}}}{{{Z}^{2}}}\] |
(4) Angular momentum | \[{{L}_{n}}=m{{v}_{n}}{{r}_{n}}=n\,\left( \frac{h}{2\pi } \right)\] | \[{{L}_{n}}\propto n\] |
(5) Corresponding current | \[{{i}_{n}}=e{{\nu }_{n}}=\frac{m{{z}^{2}}{{e}^{5}}}{4\varepsilon _{0}^{2}{{n}^{3}}{{h}^{3}}}\] | \[{{i}_{n}}\propto \frac{{{Z}^{2}}}{{{n}^{3}}}\] |
(6) Magnetic moment | \[{{M}_{n}}={{i}_{n}}A={{i}_{n}}\left( \pi \,r_{n}^{2} \right)\] (where \[{{\mu }_{0}}=\frac{eh}{4\pi m}=\] Bohr magneton) | \[{{M}_{n}}\propto n\] |
(7) Magnetic field | \[B=\frac{{{\mu }_{0}}{{i}_{n}}}{2{{r}_{n}}}=\frac{\pi {{m}^{2}}{{z}^{3}}{{e}^{7}}{{\mu }_{0}}}{8\varepsilon _{0}^{3}{{n}^{5}}{{h}^{5}}}\]c | \[B\propto \frac{{{Z}^{3}}}{{{n}^{5}}}\] |
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