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andriy [413]
3 years ago
10

A xenon arc lamp is covered with an interference filter that only transmits light of 400-nm wavelength. When the transmitted lig

ht strikes a metal surface, a stream of electrons emerges from the metal. The interference filter is then replaced with one transmitting at 300 nm and the lamp adjusted so that the intensity of the light striking the surface is the same as it was for the 400-nm light. With the 300- nm light 1. more electrons are emitted in a given time interval. 2. the electrons which are emitted are more energetic 3. both are true. 4. both are false. Please choose from 1-4, explaining your answer.
Physics
2 answers:
zepelin [54]3 years ago
4 0

Answer:

3. both are true.

Explanation:

Energy increses with decrease in wavelenght.

For photoemission to occur, a threshold energy barrier must be broken.

Higher energy means more electrons will be emmited.

The electrons emmited will posses energy that is less than the incident energy by the value of the threshold energy.

So the higher the energy, the higher the energy possessed by the electrons.

Anna35 [415]3 years ago
4 0

Answer:

3. Both are true

Explanation:

This is because the number of electrons emitted and their energy does not depend on intensity but on the wavelength. From K.E = hc/λ - Φ where λ = wavelength, Φ = work function and hc/λ = energy of the incoming light

Electrons are emitted when hc/λ > Φ. Since the energy of the incoming light is inversely proportional to its wavelength, for the 300 nm light, there is more energy and thus the difference hc/λ - Φ is more than that of the 400 nm light thereby releasing more electrons.

Also, since the difference hc/λ - Φ is more than that of the 400 nm light the kinetic energy K.E of the electrons is also more in this instant. So the electrons emitted are more energetic.

You might be interested in
The flow of energy through living systems can be modeled as food chains. True or False
Vesna [10]

Answer:

True, the flow of energy through living systems can be modeled as food chains

Explanation:

A food chain shows how energy flows from one organism to another. In general, energy flows from the Sun to producers and then to consumers. The path is linear as

the energy present in one step is transferred to the next.

For example, grasses and scattered trees growing in a field,the grass and trees are producers that use sunlight to carry out

photosynthesis. Grasshoppers are herbivores that live in grassland ecosystems. They get energy

by eating grass and leaves.

6 0
3 years ago
What does the universe look like on very large scales?
Anestetic [448]

Answer:

it looks like dots and just black space on a large scale

Explanation:

on a large scale the universe especially our milky way looks small

hope this helps  

3 0
2 years ago
What is good deductive reasoning called
andrezito [222]

to make sure the argument is true and correct backed up by logic.

6 0
3 years ago
NEED BOTH QUESTIONED ANSWERED ASAP.
Nana76 [90]

Answer:

the answer is the red super giants

8 0
3 years ago
Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In
OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

g

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

\frac{P1}{ρg} +  \frac{V1^{2} }{2g} + Z1 =  \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3

pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes

\frac{P1}{ρg} + Z1 =  \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3

Z1 = \frac{V3^{2} }{2g} + Z3

V3 = \sqrt{2g(Z1-Z3)}

V3 = \sqrt{2 x 9.8 x (10 - 3)}

V3 = 12.53 m/s

discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53

discharge rate (Q) = 0.2005 m^{3} / s

8 0
3 years ago
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