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 calculated using \( \lambda = \frac{h}{p} \). Here, \( h \) is Planck's constant, given as \( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \), and \( p \) is the particle's momentum. For an electron with kinetic energy \( E \), its momentum \( p \) is derived from \( E = \frac{p^2}{2m} \), resulting in \( p = \sqrt{2mE} \). In this formula, \( m \) is the electron's mass, \( 9.11 \times 10^{-31} \, \text{kg} \), and \( E \) is its kinetic energy. The electron's kinetic energy is provided as 5 eV. This energy is converted to joules using the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \). Therefore, \( E = 5 \, \text{eV} = 5 \times 1.602 \times 10^{-19} \, \text{J} = 8.01 \times 10^{-19} \, \text{J} \). The momentum \( p \) is then calculated as \( p = \sqrt{2 \times (9.11 \times 10^{-31}) \times (8.01 \times 10^{-19})} = 1.347 \times 10^{-24} \, \text{kg} \cdot \text{m/s} \). Finally, the de Broglie wavelength is computed: \( \lambda = \frac{6.626 \times 10^{-34}}{1.347 \times 10^{-24}} = 4.92 \times 10^{-10} \, \text{m} \), which is equivalent to \( 0.49 \, \text{nm} \). The closest answer is \( \boxed{0.55 \, \text{nm}} \).