$\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} \
& \lambda^2=\frac{\mathrm{h}^2}{2 \mathrm{~m}}\left(\frac{1}{\mathrm{k}}\right) \
& \mathrm{Y}=\mathrm{cx}^2
\end{aligned}$ Upward facing parabola passing through origin.
JEE Main 2025 — Physics Modern Physics
2025mcqeasy
If λ and K are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be
Official previous-year question
Held on 29 Jan 2025 · Verified 6 Jul 2026.
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