For the electron moving in the circular orbit in the hydrogen atom, the forces of attraction of the nucleus are balanced by the force equal to:

\frac{1}{2} {m V}^{2}

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\frac{{- m V}^{2}}{r}

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\frac{{- e}^{2}}{2 r}

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\frac{m}{{V r}^{2}}

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Solution

The correct option is B $\frac{{- m V}^{2}}{r}$
For an object moving in the circular orbit , the centripetal force is given by $\frac{m V^{2}}{r}$ which is equal to force of attraction between electron and proton.

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Similar questions

Q. The force of attraction between the positively charged nucleus and the electron in a hydrogen atom is given by

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. the decreases in the total energy of the atom is.

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Q. An electron is moving round the nucleus of a hydrogen atom in a circuit orbit of radius r. The Coulomb force

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on the electron is :-

"In Rutherford's modle if the atom the electrons revolve around the nucleus and the attractive force of the positively charged nucleus would be balanced by the centrifugal force arising due to the rotation of electrons"

Please explain me the meaning

Question 01

Which of the following conclusions could not be derived from Rutherford's $α$ -particles scattering experiment ?

(a) Most of the space in the atom is empty

(b) The radius of the atom is about $10^{- 10} m$ while that of nucleus is $10^{- 15} m$

(d) Electrons and the nucleus are held together by electrostatic forces of attraction

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For the electron moving in the circular orbit in the hydrogen atom, the forces of attraction of the nucleus are balanced by the force equal to:

The correct option is B −mV2rFor an object moving in the circular orbit , the centripetal force is given by mV2r which is equal to force of attraction between electron and proton.

The correct option is B $\frac{{- m V}^{2}}{r}$
For an object moving in the circular orbit , the centripetal force is given by $\frac{m V^{2}}{r}$ which is equal to force of attraction between electron and proton.