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) $=6.6 \times 10^{-34} \,J s$ ]

• 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 $He ^{+} 3 r d$ orbit
$E_{3}=-13.6 \times \frac{Z^{2}}{n^{2}} e V=-13.6 \times \frac{4}{9} e V $
$=-13.6 \times \frac{4}{9} \times 1.6 \times 10^{-19} J=9.7 \times 10^{-19} J$
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} $
$v=\sqrt{\frac{2 \times 9.7 \times 10^{-19}}{9.1 \times 10^{-31}}}=1.46 \times 10^{6}\, ms ^{-1}$