Atomic Structure: Practice Problem & Solution
Consider 3rd orbit of $He^+$ (Helium) using non relativistic approach the speed of electron in this orbit will be (given $K = 9 \times 10^9$ constant $Z = 2$ and h (Planck's constant) = $6.6 \times 10^{-34} Js$): (2015)
Solution Explained:
To solve this problem, we apply the core principles of Atomic Structure. Understanding the underlying formula is key to arriving at the correct answer below:
Speed of electron in nth orbit is $v_n = 2.18 \times 10^6 \frac{Z}{n} m/s$. For $He^+$, $Z=2$ and $n=3$. So, $v_n = 2.18 \times 10^6 \times \frac{2}{3} = 1.453 \times 10^6 m/s \approx 1.46 \times 10^6 m/s$.
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