Chapter 38: Q9Q (page 1180)

(a) If you double the kinetic energy of a nonrelativistic particle, how does its de Broglie wavelength change? (b) What if you double the speed of the particle?

Short Answer

Expert verified

a) The wavelength decreases by a factor of $\frac{1}{\sqrt{2}}$ .

b) The wavelength is decreases by a factor $\frac{1}{2}$ .

Step by step solution

Describe the de Broglie wavelength

The de Broglie wavelength is given by,

$λ = \frac{h}{p}$

Here, h is the Planck’s constant, and p is the momentum of the particle.

Find the change in de Broglie wavelength when the kinetic energy is doubled

(a)

The momentum of the particle and kinetic energy are related as follows.

$p = \sqrt{2 m E}$

It is known that $λ = \frac{h}{\sqrt{2 m E}}$ therefore,

$λ \propto \frac{1}{\sqrt{E}} \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{E_{2}}{E_{1}}}$

Here, $E_{2} = 2 E_{1}$ and $λ_{1} = λ$ .

$λ_{2} = λ_{1} \sqrt{\frac{E_{1}}{E_{2}}} = λ \sqrt{\frac{E_{1}}{2 E_{1}}} = \frac{λ}{\sqrt{2}}$
Therefore, the wavelength decreases by a factor of $\frac{1}{\sqrt{2}}$ .

Find the change in de Broglie wavelength when the speed of the particle is doubled

(b)

The momentum of the particle is given by,

$p = m v$

Here, v is velocity of the particle.

The wavelength of the particle is given by,

$λ = \frac{h}{m v} λ \propto \frac{1}{v} \frac{λ_{1}}{λ_{2}} = \frac{v_{2}}{v_{1}}$

Here, $v_{2} = 2 v_{1}$ , and $λ_{1} = λ$ .

$λ_{2} = λ_{1} \frac{v_{1}}{v_{2}} = \frac{v_{1}}{2 v_{1}} λ = \frac{λ}{2}$

Therefore, the wavelength is decreases by a factor $\frac{1}{2}$ .

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(a) If you double the kinetic energy of a nonrelativistic particle, how does its de Broglie wavelength change? (b) What if you double the speed of the particle?

Short Answer

Step by step solution

Describe the de Broglie wavelength

Find the change in de Broglie wavelength when the kinetic energy is doubled

Find the change in de Broglie wavelength when the speed of the particle is doubled

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