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DedPeter [7]
3 years ago
9

If the work function of a material is such that red light of wavelength 700 nm just barely initiates the photoelectric effect, w

hat must the maximum kinetic energy of ejected electrons be when violet light of wavelength 400 nm illuminates the material?Express your answer with the appropriate units.Kmax = J
Physics
1 answer:
lawyer [7]3 years ago
7 0

Answer: 2.13(10)^{-19} J

Explanation:

The photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.  

If the light is a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.  

<u>This is what Einstein proposed:  </u>

Light behaves like a stream of particles called photons with an energy  E

E=h.f (1)

Where:

h=6.63(10)^{-34}J.s is the Planck constant  

f is the frequency

Now, the frequency has an inverse relation with the wavelength \lambda:  

f=\frac{c}{\lambda} (2)  

Where c=3(10)^{8}m/s is the speed of light in vacuum  and \lambda=400nm=400(10)^{-9}m is the wavelength of the absorbed photons in the photoelectric effect.

Substituting (2) in (1):

E=\frac{h.c}{\lambda} (3)

So, the energy E of the incident photon must be equal to the sum of the Work function \Phi of the metal and the maximum kinetic energy K_{max} of the photoelectron:  

E=\Phi+K_{max} (4)  

Rewriting to find K_{max}:

K_{max}=E-\Phi (5)

Where \Phi is the minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and its value depends on the metal:

\Phi=h.f_{o}=\frac{h.c}{\lambda_{o}} (6)

Being \lambda_{o}=700nm=700(10)^{-9}m the threshold wavelength (the minimum wavelength needed to initiate the photoelectric effect)

Substituting (3) and (6) in (5):  

K_{max}=\frac{h.c}{\lambda}-\frac{h.c}{\lambda_{o}}

K_{max}=h.c(\frac{1}{\lambda}-\frac{1}{\lambda_{o}}) (7)

Substituting the known values:

K_{max}=(6.63(10)^{-34}J.s)(3(10)^{8}m/s)(\frac{1}{400(10)^{-9}m}-\frac{1}{700(10)^{-9}m})

K_{max}=2.13(10)^{-19} J >>>>>This is the maximum kinetic energy that ejected electrons must have when violet light illuminates the material

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(a)The acceleration of the 1,400-kg boat will be 0.425 m/sec²

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