Consider the 3^{rd} orbit of He^{+} (Helium) | vedclass

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(B) According to Bohr's theory,the velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:$v_n = v_0 	imes \fr

Vedclass · Accepted Answer

$1.46 	imes 10^6 \; m/s$

Vedclass · Answer

(B) According to Bohr's theory,the velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:$v_n = v_0 	imes \frac{Z}{n}$where $v_0 = 2.18 	imes 10^6 \; m/s$ is the velocity of an electron in the first orbit of the hydrogen atom.For $He^{+}$,$Z = 2$ and for the $3^{rd}$ orbit,$n = 3$.Substituting these values:$v_3 = (2.18 	imes 10^6) 	imes \frac{2}{3} \; m/s$$v_3 = 2.18 	imes 10^6 	imes 0.666... \; m/s$$v_3 \approx 1.453 	imes 10^6 \; m/s \approx 1.46 	imes 10^6 \; m/s$.

Consider the $3^{rd}$ orbit of $He^{+}$ (Helium) using a non-relativistic approach. The speed of the electron in this orbit will be (given $K = 9 \times 10^9 \; N \cdot m^2/C^2$,$Z = 2$,and $h = 6.6 \times 10^{-34} \; J \cdot s$).

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