Question

The wavelength of the electron in the first orbit of hydrogen atom is \(3.3 \times 10^{-10} \, \text{m}\). The kinetic energy of the electron (in J) is: (Given: \( h = 6.6 \times 10^{-34} \, \text{Js}, \, m_e = 9.0 \times 10^{-31} \, \text{kg} \))

• A. \( 3.33 \times 10^{-17} \, \text{J} \)
• B. \( 1.11 \times 10^{-18} \, \text{J} \)
• C. \( 2.22 \times 10^{-18} \, \text{J} \)
• D. \( 2.22 \times 10^{-17} \, \text{J} \)

Hint:

Use the relation \( KE = \frac{h^2}{2m\lambda^2} \) directly for kinetic energy with de-Broglie wavelength.

Answer 1

The Correct Option is C

From de-Broglie relation: \[ \lambda = \frac{h}{mv} \Rightarrow v = \frac{h}{m\lambda} \] Now, kinetic energy: \[ KE = \frac{1}{2}mv^2 = \frac{1}{2}m\left( \frac{h}{m\lambda} \right)^2 = \frac{h^2}{2m\lambda^2} \] Substitute: \[ KE = \frac{(6.6 \times 10^{-34})^2}{2 \cdot 9.0 \times 10^{-31} \cdot (3.3 \times 10^{-10})^2} \] \[ = \frac{43.56 \times 10^{-68}}{2 \cdot 9 \cdot 10^{-31} \cdot 10.89 \times 10^{-20}} \approx \frac{43.56 \times 10^{-68}}{196.02 \times 10^{-51}} \approx 2.22 \times 10^{-18} \, \text{J} \]