The de-Broglie wavelength $(\lambda)$ of a particle is related to its kinetic energy $(\mathrm{E})$ as

The de-Broglie wavelength $(\lambda)$ of a particle is related to its kinetic energy $(\mathrm{E})$ as
  1. $\lambda \propto \mathrm{E}$
  2. $\lambda \propto \mathrm{E}^{-1}$
  3. $\lambda \propto \mathrm{E}^{\frac{1}{2}}$
  4. $\lambda \propto \mathrm{E}^{-\frac{1}{2}}$

Solution

De-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $\therefore \quad \lambda \propto \mathrm{E}^{-\frac{1}{2}}$ ^

Asked in: MHT CET 2023 (14 May Shift 1)

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