Question

An electron is accelerated from rest through a potential difference of 200 V. The de Broglie wavelength associated with this electron is _____ nm. (Rounded off to 2 decimal places) (Planck’s constant = $6.6 × 10^{-34} J s, 1eV=1.6 × 10^{-19} J$, mass of an electron = $9.1×10^{-31} kg$)

Answer 1

Correct Answer: 0.08

An electron is accelerated from rest through a potential difference of 200 V. To find the de Broglie wavelength associated with this electron, we apply the following steps: 

1. Kinetic Energy (KE): When an electron is accelerated through a potential difference \(V\), it gains kinetic energy equal to the work done by the electric field: 
\(KE = eV\), where \(e = 1.6 \times 10^{-19} \text{ J} \text{ (charge of electron)}\).

2. Calculating KE: Here, \(V = 200 \text{ V}\), so:
\(KE = 1.6 \times 10^{-19} \times 200 = 3.2 \times 10^{-17} \text{ J}\).

3. Relating KE to momentum: The kinetic energy of the electron also relates to its momentum \(p\) by the formula: 
\(KE = \frac{p^2}{2m}\), where \(m = 9.1 \times 10^{-31} \text{ kg} \text{ (mass of electron)}\).

4. Solving for momentum (p): By rearranging, \(p = \sqrt{2m \cdot KE}\):
\(p = \sqrt{2 \times 9.1 \times 10^{-31} \times 3.2 \times 10^{-17}}\)
\(p = \sqrt{5.824 \times 10^{-47}}\)
\(p = 7.63 \times 10^{-24} \text{ kg m/s}\).

5. Applying de Broglie wavelength formula: The de Broglie wavelength \(\lambda\) is given by: 
\(\lambda = \frac{h}{p}\), where \(h = 6.6 \times 10^{-34} \text{ J s}\) (Planck's constant).

6. Calculating wavelength: \(\lambda = \frac{6.6 \times 10^{-34}}{7.63 \times 10^{-24}}\)
\(\lambda = 8.65 \times 10^{-11} \text{ m}\).

7. Converting meters to nanometers: \(1 \text{ nm} = 10^{-9} \text{ m}\), so:
\(\lambda = 0.0865 \text{ nm}\)

8. Rounding: The wavelength is \(0.09 \text{ nm}\) to 2 decimal places.

Verification: The calculated de Broglie wavelength \(0.09 \text{ nm}\).