Question

Consider $3^{rd}$ orbit of $H{e}^{+}$ (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) = $6.6\times {10}^{-34}$ J s]

• A. 3.0 × 10⁸ m/s
• B. 2.92 × 10⁶ m/s
• C. 1.46 × 10⁶ m/s
• D. 0.73 × 10⁶ m/s

Answer 1

The correct option is C

1.46 × 10⁶ m/s

$v=\frac{1}{4\pi {\epsilon }_{\circ }}\frac{2\pi z{\epsilon }^2}{nh}$

substituting the values, we get

$v=1.46\times {10}^{6}m/s$