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Chapter 38: Q 3 (page 1114)

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?

Short Answer

Expert verified

Wavelength of the light is 222.35 nm

Step by step solution

01

Step 1. It is given that electrons emerge from an aluminum surface with a maximum kinetic energy of 1.30 eV 

We need to calculate the wavelength of light.

02

Step 2. Using the expression of maximum kinetic energy.

Kmax=hf-E0f=cλKmax=hcλ-E0hcλ=Kmax+E0λ=hcKmax+E0λ=6.63×10-34×3×108m/s1.3×1.6×10-19+4.28×1.6×10-19λ=222.35nm

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Most popular questions from this chapter

a. What wavelength photon does a hydrogen atom emit in a 200→199transition?

b. What is the difference in the wavelengths emitted in a localid="1650823513478" 199→2transition and a 200→2transition?

II Ultraviolet light with a wavelength of} 70.0nm shines on a gas of hydrogen atoms in their ground states. Some of the atoms are ionized by the light. What is the kinetic energy of the electrons that are freed in this process?

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.

The electron interference pattern of Figure 38.12 was made by shooting electrons with 50keVof kinetic energy through two slits spaced role="math" localid="1650737433408" 1.0μmapart. The fringes were recorded on a detector 1.0mbehind the slits.

a. What was the speed of the electrons? (The speed is large enough to justify using relativity, but for simplicity do this as a nonrelativistic calculation.)

b. Figure 38.12 is greatly magnified. What was the actual spacing on the detector between adjacent bright fringes?

I The allowed energies of a simple atom are 0.00eV,4.00eV,and6.00eV.

a. Draw the atom's energy-level diagram. Label each level with the energy and the quantum number.

b. What wavelengths appear in the atom's emission spectrum?

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