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Consider third orbit of He $^+$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be $[K = 9 \times 10^9, Z = 2$ and $h = 6.63 \times 10^{-34} Js]$

medium

Atoms

2015

physics

$1.46 \times 10^6$ ms $^{-1}$

$0.73 \times 10^6$

Explanation

To find the speed of the electron in the third orbit of $\mathrm{He}^+$ , we use the Bohr model of the atom. $\\$ In the Bohr model, the speed of an electron in the $n$ -th orbit is given by: $\\ v_n = \frac{Z e^2}{2 \varepsilon_0 h} \cdot \frac{1}{n} \\$

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Consider third orbit of He+^+ + (Helium), using non-relativistic approach, the speed of electron in this orbit will be [K=9×109,Z=2 [K = 9 \times 10^9, Z = 2 [K=9×109,Z=2 and h=6.63×10−34Js] h = 6.63 \times 10^{-34} Js]h=6.63×10−34Js]

Explanation

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Consider third orbit of He $^+$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be $[K = 9 \times 10^9, Z = 2$ and $h = 6.63 \times 10^{-34} Js]$