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Consider 3rd orbit of $He^+$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = $9\times 10^9$ constant, Z = 2 and h (Planck's constant) = $6.6\times 10^{-34}$ Js]
A
$2.92\times 10^6$ m/s
B
$1.46\times 10^6$ m/s
C
$0.73\times 10^6$ m/s
D
$3.0\times 10^8$ m/s
Detailed Solution
$v_n = \frac{2\pi KZe^2}{nh} = \left(\frac{Z}{n}\right)\frac{2\pi Ke^2}{h}$
$\frac{2\pi Ke^2}{h} = \frac{2\times 3.14\times 9\times10^9\times(1.6\times10^{-19})^2}{6.6\times10^{-34}} \approx 2.2\times 10^6$ m/s
$v_3 = \frac{2}{3}\times 2.2\times 10^6 = 1.46\times 10^6$ m/s
$\frac{2\pi Ke^2}{h} = \frac{2\times 3.14\times 9\times10^9\times(1.6\times10^{-19})^2}{6.6\times10^{-34}} \approx 2.2\times 10^6$ m/s
$v_3 = \frac{2}{3}\times 2.2\times 10^6 = 1.46\times 10^6$ m/s