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3 years ago

List four conditions that may result in the emission of electrons from a conductor

Physics

2 answers:

QveST [7]3 years ago

8 0

There are 4 ways in which electrons are emitted from the conductor.

i. Thermionic emission

ii. Electric field electron emission

iii. Photoelectric emission

iv. Secondary emission

In thermionic emission large amount of external energy in the form of heat is supplied to release free electrons from the metal.

In electric field electron emission, electrons are emitted from the metal surface when the metals are placed in a very strong electric field.

During photoelectric emission, light is absorbed by the metals and this provides energy to the valence electrons which break their bond with the parent atom and which are then released from the atom.

Valence electrons do have some kinetic energy, but they don't have enough energy to escape from the atom. During secondary emission, a high-speed electron is bombarded with an atom, which provides the energy for the valence electrons to break their bonds with their parent atom which are then released from the atom.

NNADVOKAT [17]3 years ago

3 0

Answer:

Conditions that result in the emission of electrons from a conductor:

Heating the conductor to a suitable temperature

Exposing the conductor to a strong light

Subjecting the conductor to a very high applied voltage

Subjecting the conductor to high-speed electrons from another source

Explanation:

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hoa [83]

Here are the observations

Sugar:-

Sugar is soluble in water
so It will dissolve in water .

Corn syrup:-

Corn syrup is also basically a sugar.
It will dissolve in water too .
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3 years ago

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

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vagabundo [1.1K]

Answer:

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