Question

An electron is accelerated through a potential difference of \(10,000 V\). Its de Broglie wavelength is, (nearly): \((m_e=9\times10^{-31} kg) \)

• A. \(12.2 \times 10^{-13} m \)
• B. \(12.2 \times 10^{-12} m \)
• C. \(12.2 \times 10^{-14} m \)
• D. \(12.2\ nm\)

Answer 1

The Correct Option is B

To find the de Broglie wavelength of an electron accelerated through a potential difference, we use the formula for de Broglie wavelength given by:

\[\lambda = \frac{h}{p}\], where h is Planck's constant and p is the momentum of the electron.

The momentum p can be related to kinetic energy K.E. as:

p = \sqrt{2m_e \cdot K.E.}

The kinetic energy acquired by the electron when accelerated through a potential difference V is given by:

K.E. = eV, where e is the elementary charge.

Substitute for p in the de Broglie equation:

\[\lambda = \frac{h}{\sqrt{2m_e \cdot eV}}\]

Given:

• V = 10,000 \space V
• m_e = 9 \times 10^{-31} \space kg
• h = 6.626 \times 10^{-34} \space Js (Planck's constant)
• e = 1.602 \times 10^{-19} \space C (Elementary charge)

Substituting these values, we get:

\[\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 1.602 \times 10^{-19} \times 10,000}}\]

Calculate the value inside the square root:

2 \times 9 \times 10^{-31} \times 1.602 \times 10^{-19} \times 10,000 = 2.8836 \times 10^{-15}

Then:

\[\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2.8836 \times 10^{-15}}}\]

\[\lambda = \frac{6.626 \times 10^{-34}}{1.698 \times 10^{-7}}\]

Finally, calculate \lambda:

\[\lambda \approx 12.2 \times 10^{-12} \space m\]

Therefore, the de Broglie wavelength is approximately 12.2 \times 10^{-12} \space m, which matches the correct option.