Question

If the de Broglie wavelength of an electron is \( 2 \, \text{nm} \), then its kinetic energy is nearly (Planck's constant \( = 6.6 \times 10^{-34} \, \text{J s} \) and mass of electron \( = 9 \times 10^{-31} \, \text{kg} \))

• A. \( 0.48 \, \text{eV} \)
• B. \( 0.68 \, \text{eV} \)
• C. \( 0.38 \, \text{eV} \)
• D. \( 0.25 \, \text{eV} \)

Hint:

The de Broglie wavelength equation (\( \lambda = h/p \)) is a cornerstone of quantum mechanics, linking wave-like properties to particle momentum. Kinetic energy can be expressed in terms of momentum as \( KE = p^2/(2m) \). For calculations, ensure all units are consistent (e.g., SI units). When converting energy from Joules to electron volts, remember that \( 1 eV} = 1.6 \times 10^{-19} J} \). This conversion is frequently needed in atomic and particle physics problems.

Answer 1

The Correct Option is C

Step 1: Relate de Broglie wavelength to momentum.
The de Broglie wavelength (\( \lambda \)) is given by:
\( \lambda = \frac{h}{p} \)
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.
So, momentum \( p = \frac{h}{\lambda} \).
Step 2: Relate kinetic energy to momentum.
The kinetic energy (\( KE \)) of a particle is given by:
\( KE = \frac{1}{2}mv^2 \)
We know that momentum \( p = mv \), so \( v = \frac{p}{m} \).
Substituting \( v \) into the kinetic energy equation:
\( KE = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{1}{2}m\frac{p^2}{m^2} = \frac{p^2}{2m} \).
Step 3: Substitute the de Broglie relationship into the kinetic energy equation.
Substitute \( p = \frac{h}{\lambda} \) into the kinetic energy equation:
\( KE = \frac{\left(\frac{h}{\lambda}\right)^2}{2m} = \frac{h^2}{2m\lambda^2} \).
Step 4: Substitute the given values and calculate the kinetic energy in Joules.
Given:
\( \lambda = 2 \, \text{nm} = 2 \times 10^{-9} \, \text{m} \)
\( h = 6.6 \times 10^{-34} \, \text{J s} \)
\( m = 9 \times 10^{-31} \, \text{kg} \)
\( KE = \frac{(6.6 \times 10^{-34})^2}{2 \times (9 \times 10^{-31}) \times (2 \times 10^{-9})^2} \)
\( KE = \frac{43.56 \times 10^{-68}}{2 \times 9 \times 10^{-31} \times 4 \times 10^{-18}} \)
\( KE = \frac{43.56 \times 10^{-68}}{72 \times 10^{-49}} \)
\( KE = \frac{43.56}{72} \times 10^{-68 + 49} \)
\( KE = 0.605 \times 10^{-19} \, \text{J} \).
Step 5: Convert the kinetic energy from Joules to electron volts (eV).
We know that \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).
So, \( 1 \, \text{J} = \frac{1}{1.6 \times 10^{-19}} \, \text{eV} \).
\( KE_{\text{eV}} = \frac{0.605 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \)
\( KE_{\text{eV}} = \frac{0.605}{1.6} \)
\( KE_{\text{eV}} \approx 0.378 \, \text{eV} \).
Rounding to two decimal places, \( KE \approx 0.38 \, \text{eV} \).