Calculate de Broglie wavelength of electron accelerated through 100V

Question

Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 100 V.

(Given: mass of electron $m_e = 9.1 \times 10^{-31}$ kg, charge $e = 1.6 \times 10^{-19}$ C, $h = 6.63 \times 10^{-34}$ J·s)

Solution — Step by Step

When an electron (charge $e$ ) is accelerated through potential difference $V$ , it gains kinetic energy equal to:

KE = eV = (1.6 \times 10^{-19})(100) = 1.6 \times 10^{-17}\text{ J}

The electric field does work on the electron, converting electrical potential energy to kinetic energy.

Kinetic energy $= \frac{p^2}{2m}$ , so:

p = \sqrt{2m_e \cdot KE} = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}}

= \sqrt{2.912 \times 10^{-47}} = \sqrt{29.12 \times 10^{-48}}

= 5.40 \times 10^{-24}\text{ kg·m/s}

The de Broglie wavelength is:

\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.40 \times 10^{-24}}

= 1.23 \times 10^{-10}\text{ m} = \mathbf{1.23\text{ Å}}

The de Broglie wavelength of the electron is approximately 1.23 Å (angstroms) — comparable to atomic spacings in crystals, which is why electron diffraction is used to study crystal structures.

Why This Works

Louis de Broglie proposed in 1924 that all matter has wave-like properties, with wavelength $\lambda = h/p$ . For macroscopic objects, the wavelength is unimaginably small (a cricket ball at 30 m/s has $\lambda \sim 10^{-34}$ m). For electrons, the wavelength is in the Angstrom range — exactly the scale of atoms.

This is why electron microscopes (which use electron beams) can resolve individual atoms while optical microscopes (which use visible light, $\lambda \sim 500$ nm) cannot.

Alternative Method

A quick formula for electrons specifically: substitute $eV = \frac{p^2}{2m}$ into $\lambda = h/p$ :

\lambda = \frac{h}{\sqrt{2m_e eV}}

For $V$ in volts, this simplifies to:

\lambda = \frac{12.27}{\sqrt{V}}\text{ Å}

For $V = 100$ V: $\lambda = 12.27/\sqrt{100} = 12.27/10 = 1.227\text{ Å} \approx 1.23\text{ Å}$ .

Memorise this formula for JEE — it’s much faster for numerical problems.

The shortcut formula $\lambda = 12.27/\sqrt{V}$ Å (for electrons) is one of the most frequently used formulas in dual nature of matter numericals. JEE Main has tested this almost every year — the calculation takes 10 seconds if you remember the formula. For CBSE, derive it from first principles as shown in the step-by-step solution above.

Common Mistake

Students sometimes use $KE = \frac{1}{2}mv^2$ to find $v$ first, then calculate $p = mv$ , and finally $\lambda = h/p$ . This works but involves an extra step and risks arithmetic errors (especially with large exponents). The direct route through $p = \sqrt{2mKE}$ is faster and less error-prone. Also, never use relativistic formulas for electrons in these problems — at 100 V, the electron velocity ( $v \approx 6 \times 10^6$ m/s) is about 2% of $c$ , making relativistic corrections negligible.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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