Physics · 2015
NEET 2015 · 3 May · Code E · Q42
Consider 3^rd orbit of He^+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K =9 × 10^9 constant, Z =2…
Consider $\displaystyle 3^{\text {rd }}$ orbit of $\displaystyle \mathrm{He}^{+}$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given $\displaystyle \mathrm{K}=9 \times 10^9$ constant, $\displaystyle \mathrm{Z}=2$ and h(Planck's Constant) $\displaystyle =6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ ]
Official answer
(2)
$\displaystyle 1.46 \times 10^6 \mathrm{~m} / \mathrm{s}$
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NEET (UG) 2015 Physics question, with the answer from NTA’s final answer key. Where our answers come from.