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Chapter 38: Q 8 (page 1113)

An electron and a proton are each accelerated from rest through a potential difference of 100 V. Afterward, which particle has the larger de Broglie wavelength? Explain.

Short Answer

Expert verified

Electron will have a Higher de-broglie wavelength.

Step by step solution

01

Step 1. Given information is :An electron and a proton accelerated from rest through a potential difference of 100 V.

We need to find out the particle having the larger de Broglie wavelength.

02

Step 2. Using the formula of Wavelength and De-Broglie wavelength.

We know that :

谓=2Kmand位=hmv鈬�位=hm2Km=h2mKAsde-brogliewavelengthisinverselyproportionaltomassoftheparticle,Theparticlewithsmallermasswillhaveahigherwavelength

As the mass of electron is smaller than that of proton, Electron will have a Higher de-broglie wavelength.

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Most popular questions from this chapter

Electrons in a photoelectric-effect experiment emerge from an aluminum surface with a maximum kinetic energy of 1.30 eV. What is the wavelength of the light?

a. Calculate the de Broglie wavelength of the electron in the n=1,2,and3states of the hydrogen atom. Use the information in Table 38.2 .

b. Show numerically that the circumference of the orbit for each of these stationary states is exactly equal to n de Broglie wavelengths.

c. Sketch the de Broglie standing wave for then=3 orbit

Photoelectrons are observed when a metal is illuminated by light with a wavelength less than 388 nm. What is the metal鈥檚 work function?

a. Determine the energy, in eV, of a photon with a 550 nm wavelength.

b. Determine the wavelength of a 7.5 keV x-ray photon.

Which metals in Table 38.1 exhibit the photoelectric effect for (a) light with位=400nm and (b) light with 位=250nm?
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