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3 years ago

‏Explain how an electron emitted by the photoelectric effect can have kinetic energy less than threshold energy ?

Physics

1 answer:

xenn [34]3 years ago

4 0

Answer:

the photons (quanta of light) collide with the electrons, these electrons have to overcome the threshold energy that is the energy of union with the metal, and the energy that remains is converted to kinetic energy.

K = E - Ф

Explanation:

The photoelectric effect is the emission of electrons from the surface of a metal.

This was correctly explained by Einstein, in his explanation the energy of the photons (quanta of light) collide with the electrons, these electrons have to overcome the threshold energy that is the energy of union with the metal, and the energy that remains is converted to kinetic energy.

E = hf

E = K + Ф

K = E - Ф

The energy of the photons is given by the Planck relation E = hf and according to Einstein the number of joints must be added

E = n hf

Therefore, depending on the value of this energy, the emitted electrons can have energy from zero onwards.

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Answer:

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3 years ago

A block of gelatin is 120mm by 120mm by 40mm when unstressed. A force of 49N is applies tangentically to the upper surface causi

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3 years ago

Two point charges, q1 = 2.0 × 10−7 C and q2 = −6.0 × 10−8 C, are held 25.0 cm apart. (a) What is the electric field at a point 5

AlladinOne [14]

Answer:

a) $432000\frac{N}{C}\hat{i}$

b) $-6.92\times10^{-14}N\hat{i}$

Explanation:

The magnitude of the electric field generated by a charged particle at a distance r is:

$E= k\frac{|Q|}{r^{2}}$

With Q the charge of the particle and k the constant ()

So, the electric field generated by q1 knowing that the point 5.0 cm apart the negative charge is $25.0cm-5.0cm=20.0 cm=0.2m$ apart the positive charge is:

$E_1= k\frac{|q1|}{r_1^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|2.0\times10^{-7}|}{(0.2)^{2}}$

$E_1= 45000\frac{N}{C}$

and the electric field generated by q2:

$E_2= k\frac{|q2|}{r_2^{2}} =(9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})\frac{|-6.0\times10^{-8}|}{(0.05)^{2}}$

$E_2=216000\frac{N}{C}$

Those are the magnitudes of the electric field, but electric field is a vector quantity, so the direction is important. Electric field generated by negative particles points towards the charge and electric field generated by positive particles points away the particle. So, if we define positive direction towards negative particle (x-axis):

$\overrightarrow{E_2}=+216000\frac{N}{C}\hat{i}$

$\overrightarrow{E_1}= +45000\frac{N}{C}\hat{i}$

Vector quantities satisfy superposition principle, this is $\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}$ , with E the total electric field.

$\overrightarrow{E}=(216000+45000)\hat{i}=432000\frac{N}{C}\hat{i}$

b) The force is:

$\overrightarrow{F}=e*\overrightarrow{E}$ ,

with q the charge of an electron

$\overrightarrow{F}=(-1.61\times10^{-19})*(432000)\hat{i}=-6.92\times10^{-14}N\hat{i}$

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3 years ago

Do humans have a gravitational field if so, how strong

lilavasa [31]

I'm pretty sure that you have mass. When it comes to gravity, anything that has mass behaves exactly like anything else that has mass. Any two of them are attracted to each other, with a force that's proportional to the product of their masses, and inversely proportional to the square of the distance between them.

Now, if you have mass, then you have weight ... it's the answer you give when somebody asks "How much do you weigh ?". That's the force of gravity between you and the Earth, pulling you toward the center of the Earth. But it doesn't stop there. There's also a force of gravity pulling the Earth toward the center of you. The strength of it is EQUAL to your weight.

Your weight on Earth is EQUAL to the Earth's weight on YOU !

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3 years ago

Read 2 more answers

The uniform 12 ft pole is hinged to the truck bed and released from the vertical position as the truck starts from rest with an

Anna35 [415]

Answer:

w = 0.886 rad / s

Explanation:

Angular and linear variables are related

a = α r

where a is the linear acceleration, α the angular acceleration and r the radius of gyration

α = a / r

the angular velocity we can find it

w² = w₀² + 2 α θ

the initial angular velocity is zero, the angles to be horizontal is

θ = π/ 2 rad

we substitute

w = √ 2 a / r θ

we calculate

w = √ (2 3/12 π/2)

w = 0.886 rad / s

7 0

3 years ago