AP EAMCET · Chemistry · Structure of Atom
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
\(\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg}\right)\)
- A \(3.33 \times 10^{-17}\)
- B \(1.11 \times 10^{-18}\)
- C \(2.22 \times 10^{-18}\)
- D \(2.22 \times 10^{-17}\)
Answer & Solution
Correct Answer
(C) \(2.22 \times 10^{-18}\)
Step-by-step Solution
Detailed explanation
From de-Broglie's equation \(\mathrm{p}=\frac{\mathrm{h}}{\lambda}\) or, \(\mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda}\) K.E. \(=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~m} \cdot \frac{\mathrm{h}^2}{\mathrm{~m}^2\left(3.3 \times 10^{-10}\right)^2}\)…
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