1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
postnew [5]
3 years ago
9

If the frequency of this beam is increased while the intensity is held constant, does the number of electrons ejected per second

from the metal surface increase, decrease, or stay the same?
Physics
1 answer:
Ivan3 years ago
8 0

Answer:

if the intensity of photons is constant then number of ejected electrons will remain same

Explanation:

As per photoelectric effect we know that when light of sufficient frequency fall on the surface of metal then electrons get ejected out of the surface with certain kinetic energy

Here the energy of photons is used to eject out the electrons from metal surface and to give the kinetic energy to the ejected electrons

so we have

h\nu = W + KE

here W = work function of metal which shows the energy required to eject out electrons from metal surface

KE = kinetic energy of ejected electrons

now if we increase the frequency of the photons that incident on the metal surface then in that case the incident energy will increase

So the electrons will eject out with more kinetic energy while if the number of photon is constant or the intensity of photons is constant then number of ejected electrons will remain same

You might be interested in
A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe
Lina20 [59]

The final speed of the block after the collision with the obstacle is \boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}.

Further Explanation:

Given:

The mass of the block is 6.0\,{\text{kg}}.

The initial momentum of the block is 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}.

The impulse imparted by the obstacle is 10\,{\text{N}} \cdot {\text{s}}.

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

{p_f} - p{ & _i} = I               …… (1)                                    

Substitute 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} for {p_i} and - 10\,{\text{N}} \cdot {\text{s}} for I  in equation (1).

 \begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}

The final momentum of the block can be expressed as:

{p_f} = m{v_f}                   …… (2)                                  

Substitute 20\text{kg}\;\text{m/s} for {p_f} and 6.0\,{\text{kg}} for m in equation (2).

 \begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}

Thus, the final speed of the block after the collision with the obstacle is \boxed{3.33\;\text{m/s}}.

Learn More:

  1. Choose the 200 kg refrigerator. Set the applied force to 400 n (to the right) brainly.com/question/4033012
  2. With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward brainly.com/question/9719731
  3. Which of the following is an example of a nonpoint source of freshwater pollution brainly.com/question/1482712

Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords:  Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

5 0
3 years ago
Read 2 more answers
WILL UPVOTE EVERY ANSWER! MULTIPLE CHOICE QUESTION!
Nitella [24]
B hardness

Giddy UP!!!!!
4 0
2 years ago
Explain how the embryological development of different species reveals similarities of organisms that are not evident in the ful
denpristay [2]

Answer:Comparison of the embryological development of different species also reveals similarities that show relationships not evident in the fully-formed anatomy.

Explanation:

3 0
3 years ago
Read 2 more answers
A car moving at a speed of 36 km/h reaches the foot of a smooth
boyakko [2]

Answer:

d = 10.2 m

Explanation:

When the car travels up the inclined plane, its kinetic energy will be used to do the work in climbing up. So according to the law of conservation of energy, we can write that:

Kinetic\ Energy\ of\ the \ Car = Work\ Done\ while\ moving\ up\ the\ plane\\\frac{1}{2}mv^{2} = Fd

where,

m = mass of car

v = speed of car at the start of plane = (36 km/h)(1000 m/1 km)(1 h/3600 s)

v = 10 m/s

F = force on the car in direction of inclination = W Sin θ

W = weight of car = mg

θ = Angle of inclinition = 30°

d = distance covered up the ramp = ?

Therefore,

\frac{1}{2}mv^{2} = mgdSin\theta\\\frac{1}{2}v^{2} = gdSin\theta\\\frac{1}{2}(10\ m/s)^{2} = d(9.81\ m/s^{2}) Sin\ 30^{0}

<u>d = 10.2 m</u>

4 0
2 years ago
It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³
Evgesh-ka [11]

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, \rho_o = 800 kg/m³

density of water, \rho_w = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

8 0
3 years ago
Other questions:
  • A force of 0.2 Newtons is required to slide a book across the table. The book accelerates at 0.11 m/s squared. What is the mass
    13·1 answer
  • What is the speed of a bobsled whose distance-time graph indicates that it traveled 124m in 26s
    7·1 answer
  • A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
    11·2 answers
  • PLEASE HURRY!
    6·2 answers
  • Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi
    13·1 answer
  • What is a description of compounds?
    7·2 answers
  • Which of these correctly describes whether a girl holding a ball in the same position is doing work on the ball?
    14·1 answer
  • Oque e um projeto de vida​
    5·1 answer
  • How much of the Moon is always illuminated one time? Explain your answer.
    11·1 answer
  • Angie, brad, and carlos are discussing a physics problem in which two identical bullets are fired with equal speeds at equal-mas
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!