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Helga [31]
3 years ago
14

In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met

al are held constant. Assuming that the light incident on the metal surface causes electrons to be ejected from the metal, what happens if the intensity of the incident light is increased?
Check all that apply.

A. The work function of the metal decreases.
B. The number of electrons emitted from the metal per second increases.
C. The maximum speed of the emitted electrons increases.
D. The stopping potential increases.
Physics
1 answer:
qwelly [4]3 years ago
6 0

Answer:

B. The number of electrons emitted from the metal per second increases.

Explanation:

Light consists of photons . Energy of each photon depends upon frequency of light . The increase in intensity increases the number of photons . It does not increase energy of photons .

So if a high intensity light falls on a photosensitive plate , each photon ejects one electron . So number of electrons increases if we increase intensity of photon. It does not increase kinetic energy of ejected electrons . Work function depends upon the nature of plate.

You might be interested in
alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;
zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0
3 years ago
Which objects would sink in honey, which has a density of 1.4 g/cm³? Check all that apply.
bazaltina [42]

Answer:

any object that has density more than 1.4

Explanation:

The object that has density more than 1.4 is denser than the honey

6 0
3 years ago
Read 2 more answers
In which of the following would the particles move most rapidly? a. ice at -20 °C b. water at 20 °C c. steam at 110 °C d. boilin
Artemon [7]

Answer:

C. steam at 110

Explanation:

3 0
3 years ago
A 0.45kg baseball is pitched towards home plate at 20 m/s. The ball is hit back towards the pitcher with a speed of 30 m/s. What
Inessa05 [86]

Answer:

4.5kgm/s

Explanation:

Change in momentum is expressed as

Change in momentum = m(v-u)

M is the mass

V is the final velocity

u is the initial velocity

Given

m=0.45kg

v = 30m/s

u = 20m/s

Substitute

Change in momentum = 0.45(30-20)

Change in momentum = 0.45×10

Change in momentum = 4.5kgm/s

3 0
2 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
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