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Chapter 34: Problem 44

Electrons in a photoelectric experiment emerge from an aluminum surface with maximum kinetic energy 1.3 eV. Find the wavelength of the illuminating radiation.

Short Answer

Expert verified
The wavelength of the illuminating radiation is 954 nm

Step by step solution

01

Convert the energy from electronvolts to Joules

1.3 eV needs to be converted to Joules because Planck's constant \(h\) is in Joules. The conversion factor is \(1 eV = 1.6 x 10^{-19} J\), so \(E = 1.3 eV \times 1.6 x 10^{-19} J/eV = 2.08 x 10^{-19} J\)
02

Remember and Apply the Energy Related Formula

We use the formula \(E = hf\) to solve for \(f\). Plugging in the value for \(E\) and Planck's constant \(h = 6.626 x 10^{-34} Js\), we get \(f = E/h = (2.08 x 10^{-19} J) / (6.626 x 10^{-34} Js) = 3.14 x 10^{14} Hz\)
03

Calculate the Wavelength

Now, we know the frequency, we can use the wave equation \(f = v/λ\) where \(v = 3 x 10^8 ms^{-1}\) is the speed of light, and solve for \(λ\), the wavelength. So, \(λ = v/f = (3 x 10^8 ms^{-1}) / (3.14 x 10^{14} Hz) = 954 nm\)

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Most popular questions from this chapter

A 150 -pm X-ray photon Compton-scatters off an electron and emerges at \(135^{\circ}\) to its original direction. Find (a) the wavelength of the scattered photon and (b) the electron's kinetic energy.

Why does the photoelectric effect suggest that light has particlelike properties?

The radiance of a blackbody peaks at \(660 \mathrm{nm}\). (a) What's its temperature? (b) How does its radiance at 400 nm compare with that at \(700 \mathrm{nm} ?\)

A red laser at 650 nm and a blue laser at 450 nm emit photons at the same rate. How do their total power outputs compare?

Is it possible to determine an electron's velocity accurate to \(\pm 1 \mathrm{m} / \mathrm{s}\) while simultaneously finding its position to within \(\pm 1 \mu \mathrm{m} ?\) What about a proton?
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