The speed of an electron in the first Bohr orbit is \(7.27 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}\). The…

The speed of an electron in the first Bohr orbit is \(7.27 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}\). The speed of electron in the \(n=2\) level of \(\mathrm{Li}^{2+}\) ion will be
  1. \(7.27 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}\)
  2. \(1.09 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}\)
  3. \(1.45 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}\)
  4. \(1.82 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}\)

Solution

Since \(v \propto Z / n\), we will have \(\frac{v\left(\mathrm{Li}^{2+}ight)}{v(\mathrm{H})}=\frac{3 / 2}{1 / 1}\)
\(\Rightarrow v\left(\mathrm{Li}^{+}ight)=\left(\frac{3}{2}ight) v(\mathrm{H})=\left(\frac{3}{2}ight)\left(7.27 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}ight)=1.09 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}\) *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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