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

The question requires us to determine the de Broglie wavelength of an electron that has been accelerated through a potential difference of \(10,000 V\). We know that the de Broglie wavelength \(\lambda\) is given by the formula:

\(\lambda = \frac{h}{p}\),

where \(h\) is Planck's constant, and \(p\) is the momentum of the electron. The momentum can be found using:

\(p = \sqrt{2m_e eV}\),

where \(m_e\) is the mass of the electron, \(e\) is the charge of the electron, and \(V\) is the accelerating potential.

Given:

• \(V = 10,000 \, V\)
• \(m_e = 9 \times 10^{-31} \, kg\)
• \(e = 1.6 \times 10^{-19} \, C\)
• \(h = 6.63 \times 10^{-34} \, Js\)

Substituting these values into the momentum equation:

\(p = \sqrt{2 \times 9 \times 10^{-31} \, kg \times 1.6 \times 10^{-19} \, C \times 10,000 \, V}\)

Calculating the above expression:

\(p = \sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19} \times 10^4}\)

\(p = \sqrt{28.8 \times 10^{-16}}\)

\(p \approx 5.36 \times 10^{-24} \, kg\,m/s\)

Now, substituting for \(\lambda\):

\(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.36 \times 10^{-24}}\)

\(\lambda \approx 12.37 \times 10^{-11} \, m \approx 12.2 \times 10^{-12} \, m\)

Hence, the de Broglie wavelength of the electron is approximately \(12.2 \times 10^{-12} \, m\).

Therefore, the correct answer is \(12.2 \times 10^{-12} \, m\).