If the kinetic energy of a free electron doubles, its de-Broglie wavelength $\lambda$ changes by a factor

If the kinetic energy of a free electron doubles, its de-Broglie wavelength $\lambda$ changes by a factor
  1. $2$
  2. $\frac{1}{\sqrt{2}}$
  3. $\sqrt{2}$
  4. $\frac{1}{2}$

Solution

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2(K) m}}$ Thus, when kinetic energy $K$ is doubled, the wavelength is changes by a factor of $\frac{1}{\sqrt{2}}$.

Asked in: MHT CET 2022 (10 Aug Shift 2)

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