Question

A particle of mass \(2 \times 10^{-27} \, \text{kg}\) has de-Broglie wavelength of \(3.3 \times 10^{-10} \, \text{m}\). The kinetic energy of this particle is: (Given: Planck’s constant \( h = 6.6 \times 10^{-34} \, \text{Js} \))

• A. \( 5 \times 10^{-20} \, \text{J} \)
• B. \( 8 \times 10^{-20} \, \text{J} \)
• C. \( 1 \times 10^{-21} \, \text{J} \)
• D. \( 6 \times 10^{-22} \, \text{J} \)

Hint:

Use \( KE = \frac{h^2}{2m\lambda^2} \) directly when de-Broglie wavelength is given.

Answer 1

The Correct Option is C

de-Broglie wavelength: \[ \lambda = \frac{h}{p} = \frac{h}{mv} \Rightarrow v = \frac{h}{m\lambda} \] 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 values: \[ KE = \frac{(6.6 \times 10^{-34})^2}{2 \cdot 2 \times 10^{-27} \cdot (3.3 \times 10^{-10})^2} \] \[ = \frac{43.56 \times 10^{-68}}{2 \cdot 2 \cdot 10^{-27} \cdot 10.89 \times 10^{-20}} = \frac{43.56 \times 10^{-68}}{43.56 \times 10^{-47}} = 1 \times 10^{-21} \, \text{J} \]