Q126 / 155 · WBJEE 2025 · Physics

The de-Broglie wavelength of a moving bus with speed is $v$ is $\lambda$. Some passengers left the bus at a stoppage. Now when the bus moves with twice of its initial speed. Its kinetic energy is found to be twice of its value. What is the de-Broglie wavelength of the bus now?

  • Correct answer
    A
    $\lambda$
  • B
    $2 \lambda$
  • C
    $\frac{\lambda}{2}$
  • D
    $\frac{\lambda}{4}$

Solution

Initially $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Now acc. to question, $\mathrm{V}^{\prime}=2 \mathrm{~V}, \quad \mathrm{E}^{\prime}=2 \mathrm{E}$ $\begin{aligned} & \Rightarrow \frac{1}{2} \cdot m^{\prime}(2 \mathrm{~V})^2=2 \cdot \frac{1}{2} \mathrm{mv}^2=\mathrm{m}^{\prime}=\frac{\mathrm{m}}{2} \\ & \therefore \quad \lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{~m}}{2} \cdot 2 \mathrm{E}}}=\lambda \end{aligned}$

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

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