In a hydrogen atom,an electron of mass 9.1 ti | vedclass

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(A) The centripetal force is provided by the electrostatic force between the electron and the proton: $\frac{m_e v^2}{r} = \frac{k q^2}{r^2}$ Since $v

Vedclass · Accepted Answer

$9 	imes 10^{22} \ m \ s^{-2}, 4.1 	imes 10^{16} \ s^{-1}$

Vedclass · Answer

(A) The centripetal force is provided by the electrostatic force between the electron and the proton: $\frac{m_e v^2}{r} = \frac{k q^2}{r^2}$ Since $v = r \omega$,we can write: $m_e r \omega^2 = \frac{k q^2}{r^2} \implies \omega^2 = \frac{k q^2}{m_e r^3}$ Substituting the values: $k = 9 	imes 10^9 \ N \ m^2 C^{-2}$,$q = 1.6 	imes 10^{-19} \ C$,$m_e = 9.1 	imes 10^{-31} \ kg$,$r = 0.53 	imes 10^{-10} \ m$: $\omega = \sqrt{\frac{9 	imes 10^9 	imes (1.6 	imes 10^{-19})^2}{9.1 	imes 10^{-31} 	imes (0.53 	imes 10^{-10})^3}} \approx 4.1 	imes 10^{16} \ s^{-1}$ The radial acceleration $a_r$ is given by: $a_r = r \omega^2 = (0.53 	imes 10^{-10}) 	imes (4.1 	imes 10^{16})^2$ $a_r \approx 8.9 	imes 10^{22} \ m \ s^{-2} \approx 9 	imes 10^{22} \ m \ s^{-2}$

In a hydrogen atom,an electron of mass $9.1 \times 10^{-31} \ kg$ revolves about a proton in a circular orbit of radius $0.53 \ \mathring{A}$. The radial acceleration and angular velocity of the electron are respectively:

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