Question

If electron is accelerated under \(50 \mathrm{KV}\) in microscope, find its de- Broglie wavelength. (A) \(5.485 \times 10^{-12} \mathrm{~m}\) (B) \(8.545 \times 10^{-12} \mathrm{~m}\) (C) \(4.585 \times 10^{-12} \mathrm{~m}\) (D) \(5.845 \times 10^{-12} \mathrm{~m}\)

Short answer

The de Broglie wavelength of an electron accelerated under a \(50 \mathrm{kV}\) potential difference in a microscope is approximately \(5.485 \times 10^{-12} \mathrm{m}\), which corresponds to answer choice (A).

Step-by-step solution

Recall the de Broglie wavelength formula

The de Broglie wavelength formula is given by: \[\lambda = \frac{h}{p}\] Where \(\lambda\) is the wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the particle. For an electron accelerated under a potential difference, we can use the following formula: \[E = \frac{1}{2}mv^2 = eV\] Where: - \(E\) is the energy of the electron - \(m\) is the mass of the electron - \(v\) is the velocity of the electron - \(e\) is the elementary charge - \(V\) is the potential difference

Find the momentum

We know that the momentum is given by: \[p = mv\] Now from the energy formula, we have: \[v = \sqrt{\frac{2eV}{m}}\] Substitute the values of the energy and mass into the momentum formula: \[p = m \cdot \sqrt{\frac{2eV}{m}}\] \[p = \sqrt{2meV}\]

Calculate the de Broglie wavelength

Now, we can substitute the momentum formula into the de Broglie wavelength formula: \[\lambda = \frac{h}{\sqrt{2meV}}\] Given values: - Planck's constant (h) = \(6.626 \times 10^{-34} \mathrm{Js}\) - Elementary charge (e) = \(1.6 \times 10^{-19} \mathrm{C}\) - Mass of electron (m) = \(9.11 \times 10^{-31} \mathrm{kg}\) - Potential difference (V) = \(50 \times 10^3 \mathrm{V}\) (converted from kV to V) Substitute the given values into the wavelength formula: \[\lambda = \frac{6.626 \times 10^{-34} \mathrm{Js}}{\sqrt{2 \cdot (9.11 \times 10^{-31} \mathrm{kg}) \cdot (1.6 \times 10^{-19} \mathrm{C}) \cdot (50 \times 10^3 \mathrm{V})}}\]

Calculate the result and compare with the answer choices

Perform the calculation: \[\lambda \approx 5.485 \times 10^{-12} \mathrm{m}\] Comparing this result with the given answer choices, we find that the correct answer is (A): \[(A) \; 5.485 \times 10^{-12} \mathrm{m}\]