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: De-Broglie Wavelength Formula The de-Broglie wavelength \( \lambda \) is calculated using the following equation, which relates it to an object's momentum \( p \): \[\n\lambda = \frac{h}{p}\n\] Where: - \( h \) represents Planck's constant. - \( p \) is the object's momentum. - \( p = mv \), where \( m \) is mass and \( v \) is velocity. Therefore: \[\n\lambda = \frac{h}{mv}\n\] Step 2: Speed and Kinetic Energy Changes The bus's speed is now doubled. If the original speed is \( v \), the new speed is \( 2v \). The kinetic energy \( K \) of the bus is given by: \[\nK = \frac{1}{2} m v^2\n\] When the speed doubles, the new kinetic energy \( K' \) becomes: \[\nK' = \frac{1}{2} m (2v)^2 = 2 \times \frac{1}{2} m v^2 = 2K\n\] Thus, the new kinetic energy is twice the original value. Step 3: De-Broglie Wavelength Alteration Since the de-Broglie wavelength is inversely proportional to momentum \( p \), and momentum is directly proportional to velocity, the new de-Broglie wavelength \( \lambda' \) can be expressed as: \[\n\lambda' = \frac{h}{m(2v)} = \frac{1}{2} \times \frac{h}{mv} = \frac{\lambda}{2}\n\] Step 4: Conclusion The new de-Broglie wavelength \( \lambda' \) is half the initial wavelength \( \lambda \), so: \[\n\boxed{(C)} \, \frac{\lambda}{2}\n\]