In the Bohr model of the hydrogen atom (see Section 38.5), in the lowest energy state the electron orbits the proton at a speed of 2.2 * 106

Question

In the Bohr model of the hydrogen atom (see Section 38.5), in the lowest energy state the electron orbits the proton at a speed of 2.2 * 106 m/s in a circular orbit of radius 5.3 * 10-11 m. (a) What is the orbital period of the electron?

Verity · Answer

Answer:   Orbital period of electron is 1.5×10⁻¹⁶s   Explanation:   Given data  Proton speed v=2.2×10⁶m/s  Radius r=5.3×10⁻¹¹m  To find   Orbital period of electron T  Solution  We need to find the orbital period of electron the period is given by:  T= frac{2 pi r}{v}     Substitute the given values   T= frac{2 pi (5.3*10^{-11}m)}{2.2*10^{6}m/s }   T=1.5*10^{-16}s  Orbital period of electron is 1.5×10⁻¹⁶s

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