The speed of the electron in \( n=3 \) orbit of \( \mathrm{He^{+}} \) ion would be (given \( k = 9 \times 10^{9}\,\mathrm{N\,m^2\,C^{-2}} \); \( e = 1.6 \times 10^{-19}\,\mathrm{C} \); \( h = 6.63 \times 10^{-34}\,\mathrm{J\,s} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.46 \times 10^{6}\,\mathrm{m\,s^{-1}} \)
In Bohr's model the electrostatic force supplies the centripetal force, and the angular momentum is quantised. For an ion of atomic number \( Z \):
\( \displaystyle \frac{kZe^{2}}{r^{2}} = \frac{mv^{2}}{r} \quad\text{and}\quad mvr = \frac{nh}{2\pi} \)
Eliminating \( r \) between the two gives the speed in the \( n \)th orbit.
\( \displaystyle v = \frac{2\pi k Z e^{2}}{n h} \)
For \( \mathrm{He^{+}} \), \( Z = 2 \) and \( n = 3 \).
\( \displaystyle v = \frac{2\pi\times9\times10^{9}\times2\times(1.6\times10^{-19})^{2}}{3\times6.63\times10^{-34}} \)
\( \displaystyle v = \frac{2.895\times10^{-27}}{1.989\times10^{-33}} = 1.46\times10^{6}\,\mathrm{m\,s^{-1}} \)
Shortcut: For hydrogen, \( v_1 = 2.19\times10^{6}\,\mathrm{m\,s^{-1}} \), and \( v \propto Z/n \). So \( v = 2.19\times10^{6}\times\frac{2}{3} = 1.46\times10^{6}\,\mathrm{m\,s^{-1}} \).