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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Circular Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Circular Motion

According to the Bohr model of a hydrogen atom, the electron...

According to the Bohr model of a hydrogen atom, the electron behaves as a particle moving in a circular path centered on the proton. The proton is considered to be fixed within an inertial frame, while the required centripetal force is supplied by Coulomb attraction. For the ground state, the electron revolves around the proton in a circle of radius 5.3×10−11 m5.3 \times 10^{-11} \text{ m}. Determine the speed of the electron in this ground state. The mass of the electron is 9.1×10−31 kg9.1 \times 10^{-31} \text{ kg} and its charge is 1.6×10−19 C1.6 \times 10^{-19} \text{ C}.

Options

  1. A3.1×106 m/s3.1 \times 10^6 \text{ m/s}
  2. B2.2×106 m/s2.2 \times 10^6 \text{ m/s}
  3. C4.4×106 m/s4.4 \times 10^6 \text{ m/s}
  4. D1.1×106 m/s1.1 \times 10^6 \text{ m/s}

Correct answer

B. 2.2×106 m/s2.2 \times 10^6 \text{ m/s}

Step-by-step solution

The centripetal force acting on the electron is provided by the electrostatic force of attraction between the proton and the electron.

mv2r=14πϵ0e2r2\dfrac{mv^2}{r} = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{r^2}

v=14πϵ0e2mrv = \sqrt{\dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{mr}}

Substituting the given values:

v=9×109×(1.6×10−19)29.1×10−31×5.3×10−11v = \sqrt{\dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{9.1 \times 10^{-31} \times 5.3 \times 10^{-11}}}

v=23.04×10−2948.23×10−42v = \sqrt{\dfrac{23.04 \times 10^{-29}}{48.23 \times 10^{-42}}}

v=4.777×1012v = \sqrt{4.777 \times 10^{12}}

v≈2.18×106 m/s≈2.2×106 m/sv \approx 2.18 \times 10^6 \text{ m/s} \approx 2.2 \times 10^6 \text{ m/s}
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