Question

The kinetic energy of a fast-moving particle of mass $1 \times 10^{-31}\text{ kg}$ is associated with a de Broglie wavelength $63\text{ nm}$ is ($h = 6.3 \times 10^{-34}\text{ Js}$)}

• A. $5 \times 10^{-21}\text{ J}$
• B. $1 \times 10^{-22}\text{ J}$
• C. $5 \times 10^{-22}\text{ J}$
• D. $1 \times 10^{-21}\text{ J}$

Hint:

Always keep track of powers of 10. For $h/\lambda$, notice that $6.3/63 = 0.1$, which makes the calculation much faster.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
According to de Broglie's hypothesis, every moving particle has an associated wave. The wavelength depends on the particle's momentum, and therefore its kinetic energy.
Step 2: Key Formula or Approach:
The relation between de Broglie wavelength (\( \lambda \)) and kinetic energy (K.E) is:
\[ \lambda = \frac{h}{\sqrt{2m \cdot K.E}} \implies K.E = \frac{h^2}{2m\lambda^2} \] Step 3: Detailed Explanation:
Given: \( m = 1 \times 10^{-31} \) kg, \( \lambda = 63 \) nm = \( 63 \times 10^{-9} \) m, \( h = 6.3 \times 10^{-34} \) Js.
Substitute values into the formula:
\[ K.E = \frac{(6.3 \times 10^{-34})^2}{2 \times (1 \times 10^{-31}) \times (63 \times 10^{-9})^2} \] \[ K.E = \frac{6.3 \times 6.3 \times 10^{-68}}{2 \times 10^{-31} \times 63 \times 63 \times 10^{-18}} \] \[ K.E = \frac{(0.1 \times 63) \times (0.1 \times 63) \times 10^{-68}}{2 \times 63 \times 63 \times 10^{-49}} \] \[ K.E = \frac{0.01 \times 10^{-68}}{2 \times 10^{-49}} = \frac{10^{-70}}{2 \times 10^{-49}} \] \[ K.E = 0.5 \times 10^{-21} = 5 \times 10^{-22} \text{ J} \] Step 4: Final Answer:
The kinetic energy is \(5 \times 10^{-22}\) J.