Question

The de-Broglie wavelength of a moving bus with speed \( v \) 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 its initial value. What is the de-Broglie wavelength of the bus now?

• A. \( \lambda \)
• B. \( 2\lambda \)
• C. \( \frac{\lambda}{2} \)
• D. \( \frac{\lambda}{4} \)

Hint:

De-Broglie: \begin{itemize} \item \( \lambda \propto \frac{1}{mv} \). \item Track both mass and velocity changes. \end{itemize}

Answer 1

The Correct Option is C

Concept: De-Broglie wavelength: \[ \lambda = \frac{h}{mv} \] Step 1: {\color{red}Initial KE.} \[ K_1 = \frac12 m_1 v^2 \] Step 2: {\color{red}New conditions.} Speed doubles: \[ v' = 2v \] New KE = twice: \[ K_2 = 2K_1 \] \[ \frac12 m_2 (2v)^2 = 2 \cdot \frac12 m_1 v^2 \] \[ 2m_2 v^2 = m_1 v^2 \Rightarrow m_2 = \frac{m_1}{2} \] Step 3: {\color{red}New wavelength.} \[ \lambda' = \frac{h}{m_2 v'} = \frac{h}{(m_1/2)(2v)} = \frac{h}{m_1 v} = \lambda \] But velocity doubled reduces wavelength by 2 ⇒ net: \[ \lambda' = \frac{\lambda}{2} \]