Chapter 38, Problem 001 Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of a certain material. Photon absorption will occur if the photon energy equals or exceeds 0.42 eV, the smallest amount of energy needed to dissociate a molecule of the material.

(a) What is the greatest wavelength of light that can be absorbed by the material?
(b) In what region of the electromagnetic spectrum is this wavelength located?

Answers

Answer 1
Answer:

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=(c)/(\lambda)

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

E=h(c)/(\lambda)

And if we solve for \lambda we got:

\lambda =(hc)/(E)

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

\lambda=(4.136x10^(15)eVs (3x10^8 (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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A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student holds a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?A. 30 cmB. 60 cmC. 75 cmD. 90 cm

A 372-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 17.6 m/s2 , and its maximum speed is 1.75 m/s. a)Determine the angular frequency. b)Determine the amplitude. c)Determine the spring constant.

Answers

Answer with Explanation:

We are given that

Mass , m=372 g=(372)/(1000)=0.372 Kg

1 kg=1000g

Maximum acceleration, a=17.6 m/s^2

Maximum speed ,v=1.75 m/s

a.We know that

Maximum acceleration, a=A\omega^2

Maximum speed, v=\omega A

17.6=A\omega^2

1.75=A\omega

(17.6)/(1.75)=(A\omega^2)/(A\omega)=\omega

Angular frequency,\omega=10.06 rad/s

b.Substitute the value of angular frequency

1.75=A(10.06)

A=(1.75)/(10.06)=0.17 m

Hence, the amplitude=0.17 m

c.Spring constant,k=m\omega^2

Using the formula

k=0.372* (10.06)^2

Hence, the spring constant,k=37.6 N/m

Why is wood a renewable resource?A) We use it over and over again.
B) It takes millions of years to replace.
C) Trees grow everywhere, so we always have some.
D) It can be replaced in a reasonable amount of time.

Answers

Answer:

D) It can be replaced in a reasonable amount of time.

Explanation:

i took the test!!!!!1

Một vật chuyển động tròn đều có chu kì T = 0,25 s. Tính tần số chuyển động f của vật?

Answers

Answer:8pi

Explanatio

omega=2pi/T

Answer:

0

0000

Explanation:

Which two processes allow water to enter the atmosphere

Answers

The two processes that allow water to enter the atmosphere are: Evaporation and Transpiration.

1. **Evaporation:** Evaporation is the process by which liquid water on the Earth's surface (such as oceans, lakes, rivers, and even moist soil) is heated by the sun and turns into water vapor. This water vapor then rises into the atmosphere. Evaporation is a key component of the water cycle, where water constantly moves between the surface of the Earth and the atmosphere.

2. **Transpiration:** Transpiration is the process by which water is released into the atmosphere from plants. Plants take up water from the soil through their roots, and this water travels through the plant and eventually evaporates from small openings called stomata on the leaves. Transpiration serves various functions in plants, including cooling the plant and transporting nutrients.

Together, evaporation and transpiration contribute to the overall movement of water from the Earth's surface into the atmosphere, where it eventually condenses to form clouds and participates in various atmospheric processes before returning to the surface as precipitation through processes like rain, snow, sleet, or hail.

To know more about atmosphere:

brainly.com/question/32358340

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Answer:

sublimation and transpiration

Explanation:

The water can enter the atmosphere from snow and ice with the process of sublimation where they also make water vapors. Last water can get in the atmosphere from plants through transpiration which means that the water is evaporated through the pores of the leaves

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 4.5 m/s flies into this door at its center and bounces back at 1.5 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

Answers

Answer:

Explanation:

Mass of the gate, m_1 = 4.5 kg

Mass of the raven, m_2 = 1.2 kg

Initial speed of raven, v_1 = 4.5 m/s

Final speed of raven, v_2 = - 1.5 m/s

Moment of Inertia of the gate about the axis passing through one end:

I = (1)/(3) m_1 a^2\nI = (1)/(3) *4.5 * 2^2\nI = 6 kg m^2

Angular momentum of the gate, L = I \omega

L = 5.33 \omega

Using the law of conservation of angular momentum:

m_2 v_f (a/2) + I\omega = m_2v_i (a/2)\nI\omega = m_2 (a/2)(v_i - v_f)\n

A block m1 rests on a surface. A second block m2 sits on top of the first block. A horizontal force F applied to the bottom block pulls both blocks at constant velocity. Here m1 = m2 = m.(a)
What is the normal force exerted by the surface on the bottom block? (Use the following as necessary: m and g as necessary.)

Answers

(a) The normal force exerted by the surface on the bottom block is N1 = 2mg.

Given that,

  • A block m1 rests on a surface.
  • A second block m2 sits on top of the first block.
  • A horizontal force F applied to the bottom block pulls both blocks at constant velocity. Here m1 = m2 = m.

Based on the above information, we can say that the N1 is 2mg.

Learn more: brainly.com/question/17429689

Answer:

N = 2mg

Explanation:

Assuming the surface is horizontal

The surface must provide enough normal force to prevent the masses from accelerating in the vertical direction.