Question

The de Broglie wavelength associated with an electron moving with energy 5 eV is:

• A. 0.75 nm
• B. 1.2 nm
• C. 2.4 nm
• D. 0.55 nm

Hint:

The de Broglie wavelength of a particle is inversely proportional to its momentum. For an electron, its de Broglie wavelength depends on its energy, and a higher energy results in a shorter wavelength.

Answer 1

The Correct Option is D

The de Broglie wavelength \( \lambda \) of a particle is given by the equation: \[ \lambda = \frac{h}{p} \] where: - \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)), - \( p \) is the momentum of the particle. The momentum \( p \) of the electron can be related to its energy. For an electron with kinetic energy \( E \), the momentum is given by: \[ E = \frac{p^2}{2m} \quad \Rightarrow \quad p = \sqrt{2mE} \] where: - \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \text{kg} \)), - \( E \) is the kinetic energy of the electron. The energy of the electron is given as 5 eV. To convert this to joules, we use: \[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \] So, the kinetic energy \( E \) is: \[ E = 5 \, \text{eV} = 5 \times 1.602 \times 10^{-19} \, \text{J} = 8.01 \times 10^{-19} \, \text{J} \] Now, calculate the momentum \( p \): \[ p = \sqrt{2 \times (9.11 \times 10^{-31}) \times (8.01 \times 10^{-19})} = 1.347 \times 10^{-24} \, \text{kg} \cdot \text{m/s} \] Finally, calculate the de Broglie wavelength: \[ \lambda = \frac{6.626 \times 10^{-34}}{1.347 \times 10^{-24}} = 4.92 \times 10^{-10} \, \text{m} = 0.49 \, \text{nm} \] Thus, the closest answer is \( \boxed{0.55 \, \text{nm}} \).