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Chapter 38: Q. 17 (page 1114)

17. What is the de Broglie wavelength of a neutron that has fallen 1.0min a vacuum chamber, starting from rest?

Short Answer

Expert verified

The de Broglie wavelength of the neutron that has fallen 1.0min a vacuum chamber isλ=90nm

Step by step solution

01

Given information 

the de Broglie wavelength of a neutron that has fallen 1.0min a vacuum chamber,

02

calculation 

The neutron has traveled through the gravitational potential energy difference of

E=mgΔy

where Δy=1.0m. This potential energy difference is converted into neutron's kinetic energy, so

T=mgΔy

We also know that the kinetic energy is related to momentum through

T=p22m

Therefore

p22m=mgΔy

yielding

p=m2gΔy.

De Broglie wavelength is then simply given by

+λ-hp=hm2gΔy.

Remember that h=6.626·10-34Js, and that the mass of the neutron is m=1.67·10-27kg. With these numerical values we get

λ=90nm

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Most popular questions from this chapter

The allowed energies of a simple atom are 0.00eV, 4.00eV, and 6.00 eV. An electron traveling with a speed of 1.30×106m/scollides with the atom. Can the electron excite the atom to the n=2stationary state? The n=3stationary state? Explain.

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