N a h atom an electron revolves around a proton in a circular orbit of 0.53 armstrong calculate the period of revolution of the electron
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In the hydrogen atom, an electron revolves around a proton in a circular orbit of radius 0.53 A° . Calculate the radial acceleration and the angular velocity of the electron. me = 9.1 x 10⁻³¹ kg, e = 1.6 x 10⁻¹⁹C.
According to the coulomb's law opposite charges attract and the like charges repel with the force that is proportional to the product of the charges
Proton in a circular orbit of radius = 0.53 A = 5.3× 10`11 (Given)
e = 1.6 x 10⁻¹⁹C (Given)
The force used is the coulomb attraction force.
Therefore, the force, will be -
F =1/4πε0 ×e²/r²
where e is the charge of an electron
Thus,
F = 9 × 10`9 × (1.6 x 10⁻¹⁹)²C/ (5.3× 10`11)²
= 0.82×10`−9 N
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