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zloy xaker [14]
3 years ago
5

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:
Mumz [18]3 years ago
4 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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You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov
Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is  </em>

<em>    11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

where f_{1} is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

v = 16.27 m/s

Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound  and the reflected sound. This can be obtained as follows;

Δf = f_{2} -  f_{1}  

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf  = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

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The correct answer is:
<span>2. sound intensity is a more objective and physical attribute of a sound wave because loudness can vary from person to person

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7 0
3 years ago
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Which example describes a nonrenewable resource?
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Answer:

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Explanation:

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I = 0.50 A

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