An electron is accelerated through a potential difference of $150\text{V}$ and then passes through a double‑slit apparatus whose slit separation is $0.5\mu \text{m}$. The screen is $1\text{m}$ away from the slits. Find the distance between two consecutive bright fringes on the screen.

Step-by-step solution

1. The kinetic energy gained by the electron is $K=eV=150\times 1.6\times 10^{-19}=2.4\times 10^{-17}\text{J}$.
2. Its momentum is $p=\sqrt{2mK}$, where $m=9.11\times 10^{-31}\text{kg}$.
3. Calculate $p=\sqrt{2\times 9.11\times 10^{-31}\times 2.4\times 10^{-17}}=6.61\times 10^{-24}\text{kg⋅m/s}$.
4. The de Broglie wavelength is $\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{6.61\times 10^{-24}}=1.00\times 10^{-10}\text{m}=0.10\text{nm}$.
5. For a double slit, fringe spacing $y=\frac{\lambda L}{d}$, with $L=1\text{m}$ and $d=0.5\mu \text{m}=5\times 10^{-7}\text{m}$.
6. Compute $y=\frac{1.00\times 10^{-10}\times 1}{5\times 10^{-7}}=2.0\times 10^{-4}\text{m}=0.20\text{mm}$.

Concept tested: De Broglie wavelength and interference of matter waves

Published 7 September 2026 · How this solution was prepared and checked