Problem

An electron is accelerated through a potential difference of $50 \, \mathrm{V}$. Calculate its de Broglie wavelength.

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Data

• Potential difference: $\Delta V = 50 \, \mathrm{V}$
• Rest mass of electron: $m = 9.11 \times 10^{-31} \, \mathrm{kg}$
• Charge on electron: $q = 1.602 \times 10^{-19} \, \mathrm{C}$
• Planck’s constant: $h = 6.626 \times 10^{-34} \, \mathrm{Js}$

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Prerequisite Concepts

The de Broglie wavelength is given by:

$$\lambda = \frac{h}{\sqrt{2 q V m}}$$

where:

• $h$ is Planck’s constant,
• $q$ is the charge of the electron,
• $V$ is the accelerating potential difference,
• $m$ is the mass of the electron.

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Solution

Step 1: Substitute the Known Values

Using the formula:

$$\lambda = \frac{h}{\sqrt{2 q V m}}$$

Substitute the values:

$$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 50 \times 1.602 \times 10^{-19} \times 9.11 \times 10^{-31}}}$$

Step 2: Simplify the Denominator

Calculate the denominator step-by-step:

$$2 \times 50 \times 1.602 \times 10^{-19} \times 9.11 \times 10^{-31} = 1.459 \times 10^{-28}$$

Take the square root:

$$\sqrt{1.459 \times 10^{-28}} = 1.208 \times 10^{-14}$$

Step 3: Calculate $\lambda$

Substitute back into the formula:

$$\lambda = \frac{6.626 \times 10^{-34}}{1.208 \times 10^{-14}}$$ $$\lambda = 1.74 \times 10^{-10} \, \mathrm{m}$$

Convert to angstroms ($1 \, \mathrm{A} = 10^{-10} \, \mathrm{m}$):

$$\lambda = 1.74 \, \mathrm{\dot{A}}$$

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Answer

The de Broglie wavelength of the electron is:

$$\lambda = 1.74 \, \mathrm{\dot{A}} \, \text{or} \, 1.74 \times 10^{-10} \, \mathrm{m}.$$

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