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JEE PHYSICS MCQ: ATOMS

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In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coloumb attraction. In the ground state, the electron goes round the proton in a circle of radius 5.3×10−11 m5.3 \times 10^{-11} \mathrm{~m}. Find the speed of the electron in the ground state. Mass of the electron =9.1×10−31 kg=9.1 \times 10^{-31} \mathrm{~kg} and charge of the electron =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C}.
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Correct option: D
2.2×106 m/sec2.2 \times 10^6 \mathrm{~m} / \mathrm{sec}
Option D (2.2×106 m/sec2.2 \times 10^6 \mathrm{~m} / \mathrm{sec}) is correct because it is the calculated speed of the electron in the ground state of the hydrogen atom using Coulomb's law and centripetal force.

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