Question

Kinetic energy of a particle of mass \( 1 \times 10^{31} \, \text{kg} \) and wavelength 63 nm (where \( h = 6.3 \times 10^{-34} \, \text{Js} \))

• A. \( 1.56 \times 10^{-3} \, \text{J} \)
• B. \( 1.34 \times 10^{-3} \, \text{J} \)
• C. \( 1.00 \times 10^{-3} \, \text{J} \)
• D. \( 2.46 \times 10^{-3} \, \text{J} \)

Hint:

The de Broglie wavelength and kinetic energy are related through the momentum of the particle. Use the formula \( p = \frac{h}{\lambda} \) to find the momentum and then use the kinetic energy formula \( K.E. = \frac{p^2}{2m} \).

Answer 1

The Correct Option is B

Step 1: Use the de Broglie relation.
The de Broglie wavelength \( \lambda \) is related to the momentum \( p \) of a particle by the equation: \[ \lambda = \frac{h}{p} \] Thus, the momentum is: \[ p = \frac{h}{\lambda} \]
Step 2: Use the kinetic energy formula.
The kinetic energy \( K.E. \) of a particle is given by: \[ K.E. = \frac{p^2}{2m} \] Substituting \( p = \frac{h}{\lambda} \) into the formula: \[ K.E. = \frac{\left( \frac{h}{\lambda} \right)^2}{2m} = \frac{h^2}{2m \lambda^2} \]
Step 3: Substitute the given values.
Given: \[ h = 6.3 \times 10^{-34} \, \text{Js}, \quad m = 1 \times 10^{31} \, \text{kg}, \quad \lambda = 63 \, \text{nm} = 63 \times 10^{-9} \, \text{m} \] Substitute into the equation: \[ K.E. = \frac{(6.3 \times 10^{-34})^2}{2 \times (1 \times 10^{31}) \times (63 \times 10^{-9})^2} \]
Step 4: Calculate the value.
After calculating, the kinetic energy is: \[ K.E. = 1.34 \times 10^{-3} \, \text{J} \]
Final Answer: \( 1.34 \times 10^{-3} \, \text{J} \).