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emmainna [20.7K]
3 years ago
5

g he work functions of gold, aluminum, and cesium are 5.1 eV, 4.1 eV, and 2.1 eV, respectively. If light of a particular frequen

cy causes photoelectrons to be emitted when the light is incident on an aluminum surface, explain if we know whether this means that photoelectrons are emitted from a gold surface or a cesium surface when the light is incident on those surfaces.
Physics
1 answer:
faust18 [17]3 years ago
3 0

Answer:

Explanation:

Work function of aluminium  is  4.1 eV .

Therefor light of minimum energy of 4.1 eV will be required to cause the emission of electrons.

If we use this frequency of light or energy of light  to effect ejection of photoelectrons from cesium , it will be easily   ejected because energy requirement for this metal is lower ( work function is lower 2.1 eV ). than the energy of photon falling on it.

If we use this frequency of light or energy of light  to effect ejection of photoelectrons from aluminium  , it will be not  ejected because energy requirement for this metal is high ( work function is higher 5.1 eV ) . Even if we increase the intensity of light , electrons will not be ejected because energy of each photon depends upon the frequency .

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nika2105 [10]

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<h3>Which factors affect the critical angle for a given pair of media?</h3>

The factors which affect the critical angle are

(a) The colour (or wavelength) of light

(b) The temperature

(i) Effect of colour of light: The critical angle for a pair of media is less for the violet light and more for the red light. Thus the critical angle increases with the increase in wavelength of light.

(ii) Effect of temperature: The critical angle increases with increase in temperature because on increasing temperature of medium, its refractive index decreases.

According to the question,

μ 1​ sinCR​ =1

μ 2​ sinCY =1

μ 3​ sinCB​ =1

μ 1​ > μ 2​  and μ 2​ > μ 3

⟹μ 1​ > μ 2​ > μ 3

CR​ < CY​ < CB​

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7 0
2 years ago
A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and
notsponge [240]

Answer: 1) 17.65 * 10^-12 C/V; 2) 0.68 V

Explanation: In order to calculate the capacitance of one cylinder capacitor we have to use the following expression:

C=\frac{2\pi \epsilon o L}{ln (b/a)} where and b are the inner and outer radius of teh cylinder, respectively. L is length of the cylinder.

Finally we also kwn that C=Q/ΔV

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3 years ago
What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 7400 N
bezimeni [28]

Answer: a = 1.32 * 10^18m/s² due north

Explanation: The magnitude of the force required to move the electron is given as

F = ma

The force exerted on the charge by the electric field of intensity (E) is given by

F = Eq

Thus

Eq = ma

a = E * q/ m

Where a = acceleration of charge

E = strength of electric field = 7400N/c

q = magnitude of electronic charge = 1.609 * 10^-6c

m = mass of an electronic charge = 9.109 * 10^-31kg

a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31

a = 11906.6 * 10^-16 / 9.019 * 10^-31

a = 1.19 * 10^-12 / 9.019 * 10^-31

a = 0.132 * 10^19

a = 1.32 * 10^18m/s²

As stated in the question, the direction of the electric field is due north hence, the direction of it force will also be north thus making the electron experience a force due north ( according to Newton second law of motion)

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3 years ago
Examine the images of the Grand Canyon below. Notice that most of the canyon consists of layers of sedimentary rocks, but if you
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Intense temperature and pressure of regional metamorphism

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tiny-mole [99]

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4 years ago
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