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 \( \lambda \) of a particle is given by the formula:

\( \lambda = \frac{h}{mv} \)

where:

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

Substitute these values into the formula:

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

Calculate the denominator:

\( 9.11 \times 10^{-31} \times 6 \times 10^6 = 54.66 \times 10^{-25} = 5.466 \times 10^{-24} \)

Now calculate the de Broglie wavelength:

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

Thus, the de Broglie wavelength of the electron is approximately \( 1.2 \times 10^{-10} \, \text{m} \), which corresponds to the first option given.

Answer 2

Step 1: Write de Broglie wavelength formula 
\[ \lambda = \frac{h}{mv} \]

Step 2: Substitute values 
\[ \lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 6 \times 10^6} = \frac{6.626 \times 10^{-34}}{5.466 \times 10^{-24}} \approx 1.21 \times 10^{-10} \, m \]