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Maximum Kinetic Energy of Emitted Electrons

Created by @nathanedwards
 at November 23rd 2023, 10:37:19 pm.
AP_Physics_1-Questions
#quantum-mechanics
#photoelectric-effect
#work-function

Question:

A photon with a wavelength of 500 nm strikes a metal surface. The metal has a work function of 3.0 eV. What is the maximum kinetic energy of the emitted electron?

A) 1.5 eV B) 2.5 eV C) 3.0 eV D) 4.0 eV

Answer:

The maximum kinetic energy of the emitted electron can be found using the equation:

Kmax=hf−ϕ K_{max} = hf - \phi

Where: Kmax K_{max} = maximum kinetic energy of the emitted electron h h = Planck's constant = 6.626 x 10^-34 J*s f f = frequency of the photon ϕ \phi = work function of the metal

First, we can use the relationship between frequency and wavelength:

f=cλ f = \frac{c}{\lambda}

Where: c c = speed of light = 3.00 x 10^8 m/s λ \lambda = wavelength of the photon = 500 nm = 500 x 10^-9 m

f=3.00×108 m/s500×10−9 m f = \frac{3.00 \times 10^8\,m/s}{500 \times 10^{-9}\,m}
f≈6.00×1014 Hz f \approx 6.00 \times 10^{14}\,Hz

Now, we can calculate the energy of the photon using the equation:

E=hf E = hf

Where: E E = energy of the photon

E=6.626×10−34 J⋅s×6.00×1014 Hz E = 6.626 \times 10^{-34}\,J\cdot s \times 6.00 \times 10^{14}\,Hz
E≈3.98×10−19 J E \approx 3.98 \times 10^{-19}\,J

Converting the energy to electron volts (eV):

1 eV≈1.602×10−19 J 1\,eV \approx 1.602 \times 10^{-19}\,J
3.98×10−19 J×1eV1.602×10−19 J 3.98 \times 10^{-19}\,J \times \frac{1 eV}{1.602 \times 10^{-19}\,J}
E≈2.48eV E \approx 2.48 eV

Finally, we can find the maximum kinetic energy of the emitted electron:

Kmax=2.48 eV−3.0 eV K_{max} = 2.48\,eV - 3.0\,eV
Kmax≈−0.52 eV K_{max} \approx -0.52\,eV

Since kinetic energy cannot be negative, the maximum kinetic energy of the emitted electron is 0 eV. Therefore, the correct answer is not among the options provided.

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