Question

The gamma photons created during a PET scan are detected when they encounter a scintillator and produce a burst of light. This light is then turned into an electrical signal when it strikes a metal and an electron is emitted by means of the photoelectric effect. While this is not practical for actual medical applications, assume that the metal used in the detector is silver, with a work function of 4.7eV. Planck's constant is 4.14×10−15eV⋅s.a) What is the minimum frequency light must have to eject an electron from the surface?A. 1.1×10^12 HzB. 1.2×10^19 HzC. 2.3×10^15 HzD. 1.1×10^15 Hz

Answer 1

The minimum frequency to extract an electron is D) $1.14\cdot 10^{15} Hz$

Explanation:

The equation for the photoelectric effect is:

$hf = \phi + K_{max}$

where

(hf) is the energy of the incident photon, where

$h=4.14\cdot 10^{-15} eV\cdot s$ is the Planck constant

f is the photon frequency

$\phi$ is the work function of the material

$K_{max}$ is the maximum kinetic energy of the ejected electron

The minimum frequency of the light needed to extract an electron fro mthe surface of the material can be found by requiring that the maximum kinetic energy of the photoelectron is zero:

$K_{max}=0$

So we get

$hf=\phi$

For silver, the work function is

$\phi = 4.7 eV$

Solving for f, we find the minimum frequency:

$f=\frac{\phi}{h}=\frac{4.7}{4.14\cdot 10^{-15}}=1.1\cdot 10^{15} Hz$

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