Q85 / 100 · BITSAT 2024 (Memory Based Paper 1) · Physics

If the kinetic energy of a free electron doubles, it's de-Broglie wavelength changes by the factor

  • A
    $2$
  • B
    $\frac{1}{2}$
  • C
    $\sqrt{2}$
  • Correct answer
    D
    $\frac{1}{\sqrt{2}}$

Solution

de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{p}=\frac{h}{\sqrt{2 \cdot m \cdot(K \cdot E)}} \quad \therefore \lambda \propto \frac{1}{\sqrt{K \cdot E}}$ If K.E is doubled, $\lambda$ becomes $\frac{\lambda}{\sqrt{2}}$

Hence, the correct answer is option: 'D'.

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