Question

Calculate the de Broglie wavelength of an electron moving with velocity $ 6 \times 10^6 \, m/s $. (Mass of electron $ m = 9.11 \times 10^{-31} \, kg $, Planck's constant $ h = 6.626 \times 10^{-34} \, Js $)

• A. \(1.2 \times 10^{-10} \, m\)
• B. \(1.1 \times 10^{-10} \, m\)
• C. \(1.0 \times 10^{-10} \, m\)
• D. \(0.9 \times 10^{-10} \, m\)

Hint:

Tip: Carefully handle powers of ten in calculations involving quantum scales.

Answer 1

The Correct Option is A

Solution:

The de Broglie wavelength \( \lambda \) is calculated using the formula \( \lambda = \frac{h}{mv} \). The given parameters are:

• Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{Js} \)
• Electron mass, \( m = 9.11 \times 10^{-31} \, \text{kg} \)
• Electron velocity, \( v = 6 \times 10^6 \, \text{m/s} \)

Substituting these values yields:

\( \lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 6 \times 10^6} \)

The denominator is calculated as:

\( 9.11 \times 10^{-31} \times 6 \times 10^6 = 5.466 \times 10^{-24} \)

The de Broglie wavelength is then:

\( \lambda = \frac{6.626 \times 10^{-34}}{5.466 \times 10^{-24}} \approx 1.212 \times 10^{-10} \, \text{m} \)

Therefore, the de Broglie wavelength of the electron is approximately \( 1.2 \times 10^{-10} \, \text{m} \), matching the first provided option.