Question

The KE of particle of mass \(1 \times 10^{-31}\) Kg and the de Broglie wavelength 63 nm (h = \(6.3 \times 10^{-34}\))

Hint:

Remember that the de Broglie wavelength relates the particle's momentum to its wavelength, and kinetic energy is linked to momentum through the expression \( KE = \frac{p^2}{2m} \).

Answer 1

Step 1: Write the formula for de Broglie wavelength.
The de Broglie wavelength \( \lambda \) is given by the equation: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle.
Step 2: Express momentum in terms of kinetic energy.
The momentum \( p \) of a particle can also be related to its kinetic energy \( KE \) as: \[ p = \sqrt{2m \cdot KE} \] where \( m \) is the mass of the particle.
Step 3: Substitute the momentum expression in the de Broglie equation.
Substitute \( p = \sqrt{2m \cdot KE} \) into the de Broglie wavelength equation: \[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]
Step 4: Solve for \( KE \).
Rearranging the equation to solve for \( KE \): \[ KE = \frac{h^2}{2m \lambda^2} \]
Step 5: Substitute the given values.
Given that the mass of the particle is \( 1 \times 10^{-31} \, \text{Kg} \), the de Broglie wavelength is 63 nm (or \( 63 \times 10^{-9} \, \text{m} \)), and \( h = 6.3 \times 10^{-34} \, \text{J} \cdot \text{s} \), we can substitute these values into the equation: \[ KE = \frac{(6.3 \times 10^{-34})^2}{2 \times (1 \times 10^{-31}) \times (63 \times 10^{-9})^2} \] Now, calculating the value will give us the kinetic energy.