PYQ NEET- Atoms And Nuclei L-9

Question: Consider $3^{\text {rd }}$ orbit of $\mathrm{He}^{+}$(Helium), using non-relativistic approach, the speed of electron in this orbit will be [given $\mathrm{K}=9 \times 10^9$ constant, $\mathrm{Z}=2$ and $\mathrm{h}$ (Planck’s Constant) $=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ ]#

A) $0.73 \times 10^6 \mathrm{~m} / \mathrm{s}$

B) $3.0 \times 10^8 \mathrm{~m} / \mathrm{s}$

C) $2.92 \times 10^6 \mathrm{~m} / \mathrm{s}$

D) $1.46 \times 10^6 \mathrm{~m} / \mathrm{s}$

Answer: (D) $1.46 \times 10^6 \mathrm{~m} / \mathrm{s}$#

Sol:#

Energy of electron in $\mathrm{He}^{+} 3^{\text {rd }}$ orbit $$ \begin{aligned} & E_3=-13.6 \times \frac{4}{9} \mathrm{eV} \ & =-13.6 \times \frac{4}{9} \times 1.6 \times 10^{-19} \mathrm{~J} \ & =-9.7 \times 10^{-19} \mathrm{~J} \end{aligned} $$

According to Bohr’s model,

Kinetic energy of electron in the $3^{\text {rd }}$ orbit $=-E_3$ $$ \begin{aligned} & \therefore 9.7 \times 10^{-19}=\frac{1}{2} m_e v^2 \ & v=\sqrt{\frac{2 \times 9.7 \times 10^{19}}{9.1 \times 10^{-31}}}=1.46 \times 10^6 \mathrm{~m} / \mathrm{s} \end{aligned} $$