Question

The de-Broglie wavelength of a moving bus with speed \( v \) is \( \lambda \). Some passengers left the bus at a stop. Now, when the bus moves with twice of its initial speed, its kinetic energy is found to be twice of 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:

When the speed of an object is doubled, the de-Broglie wavelength is halved, as the wavelength is inversely proportional to velocity.

Answer 1

The Correct Option is C

Step 1: Formula for de-Broglie Wavelength The de-Broglie wavelength \( \lambda \) is related to the momentum \( p \) of the object by the following equation: \[ \lambda = \frac{h}{p} \] Where: - \( h \) is Planck's constant, - \( p \) is the momentum of the object, and - \( p = mv \) (for an object of mass \( m \) moving with velocity \( v \)). Thus: \[ \lambda = \frac{h}{mv} \] Step 2: Change in Speed and Kinetic Energy Now, the bus's speed is doubled. If the initial speed is \( v \), the new speed is \( 2v \). The kinetic energy \( K \) of the bus is given by: \[ K = \frac{1}{2} m v^2 \] When the speed is doubled, the new kinetic energy \( K' \) becomes: \[ K' = \frac{1}{2} m (2v)^2 = 2 \times \frac{1}{2} m v^2 = 2K \] Thus, the new kinetic energy is twice the initial value. Step 3: Change in de-Broglie Wavelength Since the de-Broglie wavelength depends inversely on the momentum \( p \), and momentum is directly proportional to velocity, we can write the new de-Broglie wavelength \( \lambda' \) as: \[ \lambda' = \frac{h}{m(2v)} = \frac{1}{2} \times \frac{h}{mv} = \frac{\lambda}{2} \] Step 4: Conclusion The new de-Broglie wavelength \( \lambda' \) is half of the initial wavelength \( \lambda \), so: \[ \boxed{(C)} \, \frac{\lambda}{2} \]