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(B) The velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $v_n = v_0 \frac{Z}{n}$,where $v_0$ is the velo

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$5.45 	imes 10^{5} 	ext{ m/s}$

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(B) The velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $v_n = v_0 \frac{Z}{n}$,where $v_0$ is the velocity of the electron in the ground state of the hydrogen atom.The value of $v_0$ is $2.18 	imes 10^{6} 	ext{ m/s}$.For a hydrogen atom ($H$-atom),the atomic number $Z = 1$. We need to find the speed in the fourth orbit,so $n = 4$.Substituting these values into the formula:$v_4 = (2.18 	imes 10^{6} 	ext{ m/s}) 	imes \frac{1}{4}$$v_4 = 0.545 	imes 10^{6} 	ext{ m/s}$$v_4 = 5.45 	imes 10^{5} 	ext{ m/s}$.Thus,the speed of the electron in the fourth orbit is $5.45 	imes 10^{5} 	ext{ m/s}$.

If the energy of an electron in the ground state is $-13.6 \text{ eV}$,find the speed of the electron in the fourth orbit of an $H$-atom.

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