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The de Broglie wavelength of a bus moving with speed \(v\) is \(\lambda\). Some passengers left the bus at a stopage. Now when the bus moves with twice its initial speed, its kinetic energy is found to be twice its initial value. What will be the de Broglie wavelength now? (a) \(\lambda\) (b) \(2 \lambda\) (c) \(\lambda / 2\) (d) \(\lambda / 4\)

Short Answer

Expert verified
The new de Broglie wavelength is \( \lambda/4 \), option (d).

Step by step solution

01

Understanding De Broglie Wavelength

The de Broglie wavelength is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum of the object. For an object with mass \( m \) moving with velocity \( v \), the momentum \( p \) is \( mv \). Therefore, \( \lambda = \frac{h}{mv} \).
02

Given Conditions and Changes

Initially, the bus has a velocity \( v \) and de Broglie wavelength \( \lambda = \frac{h}{mv} \). Passengers leave the bus, reducing its mass, and then it moves with twice its initial velocity, \( 2v \). We find that its new kinetic energy is twice the initial kinetic energy.
03

Relating Kinetic Energy to Changes

Kinetic energy \( KE = \frac{1}{2}mv^2 \). If the bus's new speed is \( 2v \) and the new kinetic energy equals twice the initial kinetic energy, set the kinetic energies equal: \( \frac{1}{2} m' (2v)^2 = 2 \times \frac{1}{2} mv^2 \). Simplifying gives \( 2m'v^2 = mv^2 \), so \( m' = \frac{m}{2} \).
04

Calculate New de Broglie Wavelength

With the new mass \( m' = \frac{m}{2} \) and new speed \( 2v \), the new de Broglie wavelength is \( \lambda' = \frac{h}{m' \times 2v} = \frac{h}{(\frac{m}{2})(2v)} = \frac{h}{mv} = \lambda/4 \).
05

Conclusion

Based on all calculations and reasoning, the new de Broglie wavelength of the bus will be \( \lambda/4 \). This corresponds to choice (d).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum
Momentum is a key concept in physics that describes the amount of motion an object has, which depends on its mass and velocity. You might think of it as a measure of inertia in motion. Formally, momentum \( p \) is calculated as \( p = mv \), where \( m \) is the mass and \( v \) is the velocity of the object.

- For the bus in the problem, initially moving with velocity \( v \), its momentum is \( mv \). After some passengers leave, reducing its mass, the momentum changes. If it moves with twice the velocity \( 2v \), but its kinetic energy remains doubly boosted rather than quadruple, the momentum equation helps us understand the reduction in mass affects this balance.
  • Before passengers leave, the momentum: \( p_1 = mv \)
  • After passengers leave, new momentum: \( p_2 = m'(2v) \)
These relations are crucial in understanding how both the mass and speed affect the momentum and ultimately the de Broglie wavelength.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It's a key factor in understanding how energy changes as an object moves faster or slower. Calculated as \( KE = \frac{1}{2} mv^2 \), kinetic energy depends on the square of the velocity and directly on the mass.

In the given problem, when the bus's speed doubles, its kinetic energy only doubles in value, not quadruples as one might initially expect because the bus has lost some mass:
  • Original kinetic energy: \( KE_1 = \frac{1}{2} mv^2 \)
  • New kinetic energy (double): \( KE_2 = \frac{1}{2} m'(2v)^2 = 2 \times KE_1 = mv^2 \)
To achieve this, the mass is halved \( m' = \frac{m}{2} \). These calculations show how changes in mass and velocity influence kinetic energy and help determine changes in the de Broglie wavelength.
Planck's Constant
Planck's constant \( h \) is a fundamental constant in physics that plays a crucial role in the quantum world. Its value, approximately \( 6.626 \times 10^{-34} \) Js, is what links the de Broglie wavelength equation to momentum: \( \lambda = \frac{h}{mv} \). The constant itself embodies the small, discrete packets of energy that particles in the quantum realm like to work with.

In this exercise, Planck's constant remains unchanged, acting as a scaling factor for connecting momentum and wavelength. When the bus's velocity doubles and mass decreases, it's \( h \) that ensures the product of mass and speed perfectly scales the wavelength accurately.
  • Original wavelength: \( \lambda = \frac{h}{mv} \)
  • New wavelength: \( \lambda' = \frac{h}{m'(2v)} \)
Planck's constant is a reminder of the consistency needed to bridge between energy, momentum, and wavelength, highlighting the quantum nature even in macroscopic objects like a bus.

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