Concept: The orbital velocity of an electron in a hydrogen atom depends on its principal quantum number \( n \). The speed is highest in the first orbit (ground state) and decreases as it moves to higher orbits.
Formula:$$ v_n = \frac{2\pi k e^2}{nh} $$
Solution:- For the first Bohr orbit of hydrogen, \( n = 1 \).
- Substitute the known constants: \( k = 9 \times 10^9 \), \( e = 1.6 \times 10^{-19} \text{ C} \), and \( h = 6.63 \times 10^{-34} \text{ Js} \).
- Solving this yields the standard velocity of an electron in the first Bohr orbit: \( v_1 \approx 2.18 \times 10^6 \text{ m/s} \) (often rounded to \( 2.19 \times 10^6 \text{ ms}^{-1} \)).
Why other options are incorrect:Options B, C, and D contain incorrect powers of 10. The speed of the electron in the ground state is a substantial fraction of the speed of light (approx \( \frac{c}{137} \)), requiring a large positive exponent, not negative or small powers.
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