The wavelength of electron in the first orbit of hydrogen atom is $3.3 \times 10^{-10} \mathrm{~m}$. The…

The wavelength of electron in the first orbit of hydrogen atom is $3.3 \times 10^{-10} \mathrm{~m}$. The kinetic energy of electron (in $\mathrm{J}$ ) is $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg}\right)$
  1. $3.33 \times 10^{-17}$
  2. $1.11 \times 10^{-18}$
  3. $2.22 \times 10^{-18}$
  4. $2.22 \times 10^{-17}$

Solution

From de-Broglie's equation $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ or, $\mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda}$ K.E. $=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~m} \cdot \frac{\mathrm{h}^2}{\mathrm{~m}^2\left(3.3 \times 10^{-10}\right)^2}$ $=\frac{1}{2 \times 9 \times 10^{-31}} \times\left(\frac{6.6 \times 10^{-34}}{3.3 \times 10^{-10}}\right)^2$ $\begin{aligned} & =\frac{2}{9} \times 10^{-17} \mathrm{~J}=0.222 \times 10^{-17} \mathrm{~J} \\ & =2.22 \times 10^{-18} \mathrm{~J}\end{aligned}$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

Practice more Structure of Atom questions on Aicharya