Question

The de Broglie wavelength associated with an electron accelerated through a potential difference of 2400 V is nearly

• A. 0.25 Å
• B. 0.50 Å
• C. 0.75 Å
• D. 0.95 Å

Hint:

Use $\lambda = \frac{12.27}{\sqrt{V}}$ Å for electrons in eV. Ensure V is in volts. For non-relativistic cases, this approximation holds. Compare with photon wavelength for context in electron microscopy.

Answer 1

The Correct Option is A

1. The de Broglie wavelength $\lambda$ for a particle is $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is momentum.
2. For an electron accelerated through potential $V$, kinetic energy $K = eV$, so $p = \sqrt{2 m_e e V}$.
3. Thus, $\lambda = \frac{h}{\sqrt{2 m_e e V}}$. The numerical formula in Ångströms is $\lambda \approx \frac{12.27}{\sqrt{V}}$, where $V$ is in volts.
4. Given $V = 2400$ V, $\sqrt{V} = \sqrt{2400} \approx 48.99$.
5. Calculate $\lambda \approx \frac{12.27}{48.99} \approx 0.25$ Å.
6. Therefore, the correct option is (1) 0.25 Å.