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Setler [38]
3 years ago
7

What do we call the minimum energy that is required by an electron to leave the metal target in the photoelectric effect?Select

one of the options below as your answer: A. energy function B. minimum function C. work function D. quanta E. electrical energy
Physics
2 answers:
Liula [17]3 years ago
7 0

Answer:

C. work function

Explanation:

In the photoelectric effect, the energy of the incident photon is used in part to extract the electron from the metal (and this energy is called work function) and the rest is converted into kinetic energy of the electron. In formula:

hf = \phi +K

where

hf is the energy of the incident photon, which is the product between h (the Planck constant) and f (the photon's frequency)

\phi is the work function

K is the kinetic energy of the photoelectron as it leaves the material

Sergio [31]3 years ago
6 0
The work function is what we call the minimum energy that is required by an electron to leave the metal target in the photoelectric effect. 
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If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
kaheart [24]

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

\Delta T is the temperature of difference that we want to find

\Delta x=2.5 cm =0.025 m represent the thickness of the material

If we solve \Delta T in absolute value from the equation (1) we got:

\Delta T =\frac{Q \Delta x}{Ak}

First we convert 3KW to W and we got:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

And we have everything to replace and we got:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

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