The electrostatic potential energy of the ele | vedclass

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(B) Given,electrostatic potential energy $U = -6.8 \ eV$.We know that for a hydrogen-like atom,the total energy $E_n$ is related to the potential ener

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$\frac{C}{274}$

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(B) Given,electrostatic potential energy $U = -6.8 \ eV$.We know that for a hydrogen-like atom,the total energy $E_n$ is related to the potential energy $U$ by the relation $E_n = \frac{U}{2}$.Therefore,$E_n = \frac{-6.8 \ eV}{2} = -3.4 \ eV$.The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = \frac{-13.6 \ eV}{n^2}$.Equating the two,we get $\frac{-13.6}{n^2} = -3.4$,which implies $n^2 = 4$,so $n = 2$.The speed of an electron in the $n^{th}$ orbit is given by $v_n = \frac{C}{137n}$.Substituting $n = 2$,we get $v_2 = \frac{C}{137 	imes 2} = \frac{C}{274}$.

The electrostatic potential energy of the electron in an orbit of hydrogen is $-6.8 \ eV$. The speed of the electron in this orbit is ($C$ is the speed of light in vacuum).

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