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By what factor does the de Broglie wavelength of a particle change if \((a)\) its momentum is doubled or \((b)\) its kinetic energy is doubled? Assume the particle is nonrelativistic.

Short Answer

Expert verified
(a) Wavelength halves ( frac{1}{2}); (b) Wavelength decreases by \( frac{1}{ sqrt{2}}\).

Step by step solution

01

Understanding the de Broglie Wavelength Formula

The de Broglie wavelength formula is given by \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle.
02

Step (a): Analyzing Momentum Doubling

If the momentum \( p \) is doubled \(( p' = 2p )\), the new wavelength \( \lambda' \) becomes \( \lambda' = \frac{h}{2p} \). This shows that the wavelength is halved when the momentum is doubled.
03

Step (a) Conclusion

The de Broglie wavelength of a particle becomes half of its original wavelength when its momentum is doubled. Therefore, the factor by which the wavelength changes is \( \frac{1}{2} \).
04

Step (b): Relating Kinetic Energy to Momentum

The kinetic energy \( KE \) of a nonrelativistic particle is given by \( KE = \frac{p^2}{2m} \), where \( m \) is the mass of the particle. If the kinetic energy is doubled, we have \( KE' = 2KE = \frac{(p')^2}{2m} \), leading to \( 2 \frac{p^2}{2m} = \frac{(p')^2}{2m} \), so \( (p')^2 = 2p^2 \). This implies \( p' = p\sqrt{2} \).
05

Step (b): Calculating the New Wavelength

With the new momentum, \( p' = p\sqrt{2} \), the new de Broglie wavelength is \( \lambda' = \frac{h}{p\sqrt{2}} \). This results in \( \lambda' = \frac{1}{\sqrt{2}} \lambda \), indicating that the wavelength decreases by a factor of \( \frac{1}{\sqrt{2}} \) when kinetic energy is doubled.
06

Step (b) Conclusion

The de Broglie wavelength of a particle decreases by a factor of \( \frac{1}{\sqrt{2}} \) when its kinetic energy is doubled. Therefore, the factor by which the wavelength changes is \( \frac{1}{\sqrt{2}} \).

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Key Concepts

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

Momentum
Momentum is a crucial concept in physics that describes the quantity of motion an object possesses. It is defined as the product of an object's mass and velocity, represented by the formula \( p = mv \). Momentum is a vector quantity, meaning it has both magnitude and direction.

When considering the de Broglie wavelength, momentum plays a direct role. The de Broglie wavelength \( \lambda \) of a particle is inversely proportional to its momentum as given by the formula \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant. So, by increasing momentum, the wavelength decreases and vice versa.

In the scenario where the particle's momentum is doubled \(( p' = 2p )\), the wavelength is halved. This is because the wavelength becomes \( \lambda' = \frac{h}{2p} \), highlighting how intimately connected momentum and the de Broglie wavelength are.
Kinetic Energy
Kinetic energy refers to the energy that an object possesses due to its motion, calculated through the formula \( KE = \frac{1}{2} mv^2 \). For nonrelativistic particles, kinetic energy can also be expressed as a function of momentum: \( KE = \frac{p^2}{2m} \), where \( m \) is the mass of the particle. This linkage allows us to see how changes in kinetic energy affect momentum.

When kinetic energy is doubled, the momentum transforms according to \( KE' = 2KE = \frac{(p')^2}{2m} \), leading to \( (p')^2 = 2p^2 \). Consequently, the new momentum \( p' \) becomes \( p\sqrt{2} \). This change in momentum then influences the de Broglie wavelength, which decreases by a factor of \( \frac{1}{\sqrt{2}} \) when kinetic energy is doubled, showing how kinetic energy impacts particle properties.
Nonrelativistic Particles
Nonrelativistic particles are those moving at speeds significantly less than the speed of light, making relativistic effects negligible. This categorization simplifies calculations because relativistic corrections (which arise at high speeds) are not required.

In terms of the de Broglie wavelength, the assumption of nonrelativistic speeds allows us to use simple classical formulas to approximate momentum and kinetic energy. This makes the relation between these quantities and the de Broglie wavelength straightforward.

For such particles, we can confidently use the formulas \( \lambda = \frac{h}{p} \) for wavelength, \( p = mv \) for momentum, and \( KE = \frac{1}{2} mv^2 \) or \( KE = \frac{p^2}{2m} \) for kinetic energy without considering relativistic mass changes. This simplicity is why nonrelativistic assumptions are usually made in fundamental quantum mechanics calculations.

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Most popular questions from this chapter

White light, with frequencies ranging from \(4.00 \times 10^{14} \mathrm{Hz}\) to \(7.90 \times 10^{14} \mathrm{Hz},\) is incident on a potassium surface. Given that the work function of potassium is \(2.24 \mathrm{eV}\), find (a) the maximum kinetic energy of electrons ejected from this surface and (b) the range of frequencies for which no electrons are ejected.

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