Wavelength Ratio For Two Accelerating Potentials

For an electron accelerated from rest through potential $V$, the de Broglie wavelength is $\lambda =\frac{h}{\sqrt{2meV}}$, so for a fixed charge and mass $\lambda \propto \frac{1}{\sqrt{V}}$. Taking the two cases, $\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{V_2}{V_1}}=\sqrt{\frac{200}{50}}=\sqrt{4}=2$. Thus the electron accelerated through the smaller potential of $50\text{V}$ has the longer wavelength, in a ratio of $2:1$. The option $4:1$ forgets the square root and uses the bare voltage ratio. The option $1:2$ inverts the relationship by placing the higher voltage in the numerator. The option $\sqrt{2}:1$ would correspond to a voltage ratio of two rather than four. The inverse-square-root dependence means a fourfold increase in accelerating voltage only halves the matter wavelength, a slow scaling that experimenters exploit to tune electron wavelengths in diffraction and microscopy. This builds directly on the accelerated-electron formula central to the NCERT dual-nature chapter. A check confirms the lower-energy electron has the longer wave, as physical intuition about momentum demands. A practical implication of this gentle inverse-square-root scaling is that achieving a substantially shorter electron wavelength demands a disproportionately large increase in accelerating voltage, which is why high-resolution electron microscopes operate at tens or hundreds of kilovolts. The slow dependence also means modest voltage instabilities translate into only small wavelength fluctuations, which helps keep diffraction patterns sharp.