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user100 [1]
3 years ago
14

Chapter 38, Problem 001

Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

a) \lambda=2.95x10^{-6}m

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

E=hf

Where h is the Planck's Constant. By the other hand the know that c=f\lambda and if we solve for f we have:

f=\frac{c}{\lambda}

If we replace the last equation into the E formula we got:

E=h\frac{c}{\lambda}

And if we solve for \lambda we got:

\lambda =\frac{hc}{E}

Using the value of the constant h=4.136x10^{-15} eVs we have this:

\lambda=\frac{4.136x10^{15}eVs (3x10^8 \frac{m}{s})}{0.42eV}=2.95x10^{-6}m

\lambda=2.95x10^{-6}m

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of 10^{-6}m

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How does laser works ?<br>​
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Explanation:

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3 years ago
Two large metal plates of area 0.7 m2 face each other. They are 5.1 cm apart and have equal but opposite charges on their inner
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Explanation:

The given data is as follows.

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Since, electric field is related to surface charge density as follows.

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Also,    E = \frac{\sigma}{\epsilon_{o}}

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Therefore, charge will be calculated as follows.

          Q = \sigma \times A = \epsilon_{o} E \times A

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If all of the forces acting on an object are balanced, then:
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A. the direction the object is moving in will not change.

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\sum F=0

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In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le
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Answer:

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Explanation:

From the question we are told that

     The  focal length of the objective is  f_o =  1.0 \ cm

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