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Physics Question on Atoms

Consider 3rd3^{rd} orbit of He+He^+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K=9×109 K= 9 \times 10^9 constant, Z = 2 and h (Planck's Constant) = 6.6×10−34Js6.6 \times 10^{-34} J s]

A

0.73 × 106\times \, 10^6m/s

B

3.0 × 108\times \, 10^8m/s

C

2.92 × 106\times \, 10^6m/s

D

1.46 × 106\times \, 10^6m/s

Answer

1.46 × 106\times \, 10^6m/s

Explanation

Solution

Energy of electron in He+^+ 3rd orbit
E3=−13.6×Z2n2eV=−13.6 ×49eVE_3=-13.6 \times \frac{Z^2}{n^2}eV=-13.6 \, \times \frac{4}{9}e V
=−13.6×49×1.6×10−19j=9.7×10−19J=-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 3rd orbit = - E3_3
∴   9.7×10−19=12mev2\therefore \, \, \, 9.7 \times 10^{-19}=\frac{1}{2}m_e v^2
v=2×9.7×10−199.1×10−31=1.46×106 ms−1v=\sqrt{\frac{2 \times 9.7 \times 10^{-19}}{9.1 \times 10^{-31}}}=1.46 \times 10^6 \, ms^{-1}