Question

Consider $3^{\text{rd}}$ 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) $=6.6\times 10^{-34}Js$ ]

• A. 0.73 $\times 10^6$m/s
• B. 3.0 $\times 10^8$m/s
• C. 2.92 $\times 10^6$m/s
• D. 1.46 $\times 10^6$m/s

Answer 1

Answer: (D)

Energy of electron in $H{e}^{+}3rd$ orbit
$E_3=-13.6\times \frac{Z^2}{n^2}eV=-13.6\times \frac{4}{9}eV$
$=-13.6\times \frac{4}{9}\times 1.6\times 10^{-19}J=9.7\times 10^{-19}J$
As per Bohr's model,
Kinetic energy of electron in the $3^{\text{rd}}$ orbit $=-E_3$
$\therefore 9.7\times 10^{-19}=\frac{1}{2}{m}_e{v}^2$
$v=\sqrt{\frac{2\times 9.7\times 10^{-19}}{9.1\times 10^{-31}}}=1.46\times 10^{6}m{s}^{-1}$