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deff fn [24]
2 years ago
13

Light of a given wavelength is used to illuminate the surface of a metal, however, no photoelectrons are emitted. in order to ca

use electrons to be ejected from the surface of this metal you should:_________
Physics
1 answer:
solniwko [45]2 years ago
8 0

We will decrease the wavelength of light in order to cause electrons to be ejected from the surface of this metal.

We have a Light of a given wavelength that is used to illuminate the surface of a metal, however, no photoelectrons are emitted.

We have to find out what can we do in order to cause electrons to be ejected from the surface of this metal.

<h3>What is Photoelectric Effect ?</h3>

The emission of electrons from the surface of the metal when the light of specific frequency (greater than the threshold frequency) falls over it is called photoelectric effect.

Light consists of photons. The energy associated with the photons is used to emit out the electrons from the surface of metal. We know that - Energy can neither be created nor be destroyed and it can only be transferred from one body to another. Hence, the energy of these moving photons is used to emit electrons from the metal surface. The energy associated with the photon is given by -

E = hμ

Where - μ is the frequency of light an h is Planck's constant.

Now, we can see that the energy of the photon is directly proportional to the frequency of light. The minimum frequency required to eject the electron from the metal surface is called Threshold frequency. Thus, we can emit the electron from the metal surface by using the light of frequency greater than threshold frequency.

Hence, we will increase the frequency of light in order to cause electrons to be ejected from the surface of this metal

To solve more questions on Photoelectric Effect, visit the link below -

brainly.com/question/9260704

#SPJ4

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Which one of the following scenarios accurately describes a condition in which resonance can occur? A. A column of air has a hei
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The following scenarios that accurately describes a condition in which resonance can occur is vibrating tuning fork is struck and begins to vibrate as the object used to strike it is placed away from the tuning fork. The answer is letter B.

3 0
4 years ago
The index of refraction for red light in water is 1.331 and that for blue light is 1.340. A ray of white light enters the water
Alex Ar [27]

Answer:

(a) 47.08°

(b) 47.50°

Explanation:

Angle of incidence  = 78.9°

<u>For blue light : </u>

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₂ is the refractive index for blue light which is 1.340

n₁ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{78.9}^0}=\frac {1}{1.340}

{sin\theta_2}=0.7323

Angle of refraction for blue light = sin⁻¹ 0.7323 = 47.08°.

<u>For red light : </u>

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₂ is the refractive index for red light which is 1.331

n₁ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{78.9}^0}=\frac {1}{1.331}

{sin\theta_2}=0.7373

Angle of refraction for red light = sin⁻¹ 0.7373 = 47.50°.

5 0
4 years ago
What acceleration is produced on a mass of 200g, when a force of 10N is exerted on it?​
n200080 [17]

Answer:

f=ma......10N=0.2a....=50m/s

4 0
3 years ago
A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and
-Dominant- [34]

Answer:

Explanation:

Check attachment for solution

4 0
3 years ago
Read 2 more answers
Young's experiment is performed with light of wavelength 502nm from excited helium atoms. Fringes are measured carefully on a sc
alexandr402 [8]

Answer:

d = 1.30 mm

Explanation:

given,

wavelength of the light source (λ)= 502 nm

slits is separated (d) = ?

distance to form interference pattern(D) = 1.35 m  

20 th fringe

y = 10.4 mm = 0.0104 m

now,separation between to slits

 d = \dfrac{m\lambda\ D}{y}

where y is the fringe width

 d = \dfrac{20 \times 502\times 10^{-9}\ \times 1.35}{0.0104}

 d = \dfrac{13354\times 10^{-9}}{0.0104}

      d = 1.30 mm

separation between to slits is equal to d = 1.30 mm

8 0
3 years ago
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