Test: Dual Nature of Radiation & Matter - 1 - NEET MCQ

# Test: Dual Nature of Radiation & Matter - 1 - NEET MCQ

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## 30 Questions MCQ Test Physics Class 12 - Test: Dual Nature of Radiation & Matter - 1

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Test: Dual Nature of Radiation & Matter - 1 - Question 1

### Work function of a metal is the

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 1

In the photoelectric effect, the work function is the minimum amount of energy (per photon) needed to eject an electron from the surface of a metal.Electrons ejected from a sodium metal surface were measured as an electric current.The minimum energy required to eject an electron from the surface is called the photoelectric work function.
This energy (work function) is a measure of how firmly a particular metal holds its electrons. The work function is important in applications involving electron emission from metals, as in photoelectric devices and cathode-ray tubes.

Test: Dual Nature of Radiation & Matter - 1 - Question 2

### Photon has a charge of

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 2

A photon is massless, has no electric charge, and is a stable particle. In vacuum, a photon has two possible polarization states. The photon is the gauge boson for electromagnetism, and therefore all other quantum numbers of the photon (such as lepton number, baryon number, and flavour quantum numbers) are zero.

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Test: Dual Nature of Radiation & Matter - 1 - Question 3

### The work function of a photoelectric material is 3.32 eV. The threshold frequency will be equal to

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 3

Threshold frequency: - frequency of minimum energy required to remove electron

work function=3.3ev
f=3.3×1.6×10−19J​ /6.626×10−34J−s
=0.8×1015Hz
=8×1014Hz

Test: Dual Nature of Radiation & Matter - 1 - Question 4

Energy of a photon of green light of wavelength 5500m  is (given: h = 6.62 ×10−34Js−1) approximately

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 4

As we know,
the formula for energy of photon in terms of wavelength is,, E = hc/ λ
where, E = energy of photon
h = Planck's constant =
= 6.63×10-34 Js
c = speed of light
= 3×108 m/s
and lambda = wavelength
so,
E = [6.63×10-34 J. s×3×108 m/s]/5500×10-10m
= [0.36×10-26]/10-8
= 0.36×10-18
= 3.6×10-19 J
Or, 2.26ev

Test: Dual Nature of Radiation & Matter - 1 - Question 5

Uniform electric and magnetic fields are produced pointing in the same direction. An electron is projected pointing in the same direction, then

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 5

Electromagnetic force on electron will be zero V×B=0
Electrostatic force will be Fe​=−E backward, hence electron decelerate and velocity will decrease in magnitude.

Test: Dual Nature of Radiation & Matter - 1 - Question 6

In which case is electron emission from a metal not known?

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 6

The correct answer would be option B. Applying a very strong magnetic field.
Applying a very strong magnetic field to a metal we don’t know the electron emission.

Test: Dual Nature of Radiation & Matter - 1 - Question 7

Photons can be

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 7

A photon of wavelength 6000 nm collides with an electron at rest. After scattering, the wavelength of the scattered photon is found to change by exactly one Compton wavelength.

Test: Dual Nature of Radiation & Matter - 1 - Question 8

Given h = 6.6 ×10−34 joule sec, the momentum of each photon in a given radiation is 3.3 ×10−29 kg metre/sec. The frequency of radiation is

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 8

Frequency=C/ λ
λ=h/P
frequency=C P/ h
f=(3×108×3.3×10-29)/6.6×10-34
f=3× 1013/2
f=1.5×1013 Hz

Test: Dual Nature of Radiation & Matter - 1 - Question 9

If the work function of a material is 2eV, then minimum frequency of light required to emit photo-electrons is

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 9

Φ= hνλ => 2eV= 6.626 x 10-34 x ν

2 x 1.6 x 10-19= 6.626 x 10-34 x ν

On solving we get ν = 4.6 x 1014 Hz.

Test: Dual Nature of Radiation & Matter - 1 - Question 10

In Thomson’s method for finding specific charge of positive rays, the electric and magnetic fields are

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 10

The electric and magnetic fields are Parallel and separated because magnetic field applies force only on moving charge and electric field applies force both moving and stationery charges.

Test: Dual Nature of Radiation & Matter - 1 - Question 11

Photoelectric effect is

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 11

Photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it. In a broader definition, the radiant energy may be infrared, visible, or ultraviolet light, X-rays, or gamma rays; the material may be a solid, liquid, or gas; and the released particles may be ions (electrically charged atoms or molecules) as well as electrons.

Test: Dual Nature of Radiation & Matter - 1 - Question 12

In a photon-particle collision (such as photon-electron collision) the quantity which is not conserved is

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 12

In a photon-electron collision both total energy and total momentum are conserved. As in the case of the Compton effect,  when a photon with some energy collides with a stationary electron, some of the energy and momentum is transferred to the electron  but both energy and momentum are conserved in this elastic collision.so the correct option would be option B.

Test: Dual Nature of Radiation & Matter - 1 - Question 13

The specific charge for positive rays is much less than that for cathode rays. This is because

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 13

Cathode rays are high energetic electron beam.
Specific charge of cathode rays = e​/me​
Specific charge of positive rays =∣e∣/m​
Since the mass of positive rays is much larger than that of cathode rays, so the specific charge for positive rays is much smaller than that for cathode rays.

Test: Dual Nature of Radiation & Matter - 1 - Question 14

Photoelectric effect is possible

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 14

Photoelectrons will come out from the surface of the metal only if it gets enough energy during the irradiation the energy needed depends on the metal and the max energy that can be provided on the light.

Test: Dual Nature of Radiation & Matter - 1 - Question 15

Davisson Germer experiment proves that

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 15

Davisson and Germer Experiment, for the first time, proved the wave nature of electrons and verified the de Broglie equation. de Broglie argued the dual nature of matter back in 1924, but it was only later that Davisson and Germer experiment verified the results. The results established the first experimental proof of quantum mechanics. In this experiment, we will study the scattering of electrons by a Ni crystal.

Test: Dual Nature of Radiation & Matter - 1 - Question 16

If the voltage across the electrodes of a cathode ray tube is 500 volts then energy gained by the electrons is

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 16

Voltage across the electrodes of a cathode ray gun, V=500V
Charge of the electron=1.6x10-19
Energy=eV
E= 1.6x10-19 x 500
E=800 x 10-19 J
E=8 x 10-17 J

Test: Dual Nature of Radiation & Matter - 1 - Question 17

If work function of a metal plate is negligible then the K.E.of the photoelectrons emitted when radiations of 1000 Â are incident on the metal surface is

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 17

Incident frequency=work function + kinetic energy.
incident frequency=E.
therefore, hc/λ=E
for energy in e.v & wavelength in Angstrom...
Now,
we know,
E=12400/λ Å
E(in e.v)=12400/1000.=12.4e.v

Test: Dual Nature of Radiation & Matter - 1 - Question 18

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential is

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 18

Given, the maximum kinetic energy: Kmax​=4eV
If V0​ be the stopping potential, then Kmax​=eV0
⇒eV0​=4eV
⇒V0​=4V

Test: Dual Nature of Radiation & Matter - 1 - Question 19

In Photoelectric effect

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 19

Photoelectric cell or photocell, device whose electrical characteristics (e.g., current, voltage, or resistance) vary when light is incident upon it. The most common type consists of two electrodes separated by a light-sensitive semiconductor material.
The photoelectric effect is the observation that many metals emit electrons when light shines upon them. Electrons emitted in this manner can be called photoelectrons.

Test: Dual Nature of Radiation & Matter - 1 - Question 20

A photon is

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 20

A particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass.

Test: Dual Nature of Radiation & Matter - 1 - Question 21

If maximum velocity with which an electron can be emitted from a photo cell is 3.75×108cms−1 then stopping potential is

Test: Dual Nature of Radiation & Matter - 1 - Question 22

In an experiment of photoelectric emission for incident light of 4000 Â, the stopping potential is 2V. If the wavelength of incident light is made 3000 Â, then stopping potential will be

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 22

The maximum kinetic energy for the photoelectrons is
Emax​=hν−ϕ
where, ν is the frequency of incident light and ϕ is photoelectric work function of metal.
If Vo​ is the stopping potential then
eV0​=h(c/λ)​−ϕ .....................(since, ν=c/λ​)
As per the problem, for incident light of 4000Ao, the stopping potential is 2V. When the wavelength of incident light is reduced to 3000Ao, then the stopping potential will increase to value more than 2V(as per the above equation).

Test: Dual Nature of Radiation & Matter - 1 - Question 23

In the above experimental set up for studying photoelectric effect, if keeping the frequency of the incident radiation and the accelerating potential fixed, the intensity of light is varied, then

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 23

The number of electrons emitted per second is observed to be directly proportional to the intensity of light. “Ok, so light is a wave and has energy. It takes electrons out of a metal, what is so special about that!” First of all, when the intensity of light is increased, we should see an increase in the photocurrent (number of photoelectrons emitted). Right?
As we see, this only happens above a specific value of frequency, known as the threshold frequency. Below this threshold frequency, the intensity of light has no effect on the photocurrent! In fact, there is no photocurrent at all, however high the intensity of light is.

The graph between the photoelectric current and the intensity of light is a straight line when the frequency of light used is above a specific minimum threshold value.

Test: Dual Nature of Radiation & Matter - 1 - Question 24

Photo-electric effect can be explained only by assuming that light

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 24

Photoelectric effect can be explained by assuming light consisting of quanta that when a photon of certain energy falls on the surface of metal, then the metal surface starts to emit an electron known as photo electrons.

Test: Dual Nature of Radiation & Matter - 1 - Question 25

Wavelength of light incident on a photo cell is 3000 Â, if stopping potential is 2.5 volts, then work function of the cathode of photo cell is

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 25

The Stopping potential =2.5V.
or, Kinetic energy=2.5eV.
We know that,
Incident energy =work function + Kinetic energy.
To get incident energy in e.V,
We also know that,
The Incident energy =12400/λ Å
Incident energy=work function + kinetic energy.
12400/3000 = work function + 2.5e.v.
4.13-2.5 = work function
work function=1.64 e.V

Test: Dual Nature of Radiation & Matter - 1 - Question 26

in photoelectric effect, the photoelectric current

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 26

When intensity of incident photons increases, the number of electrons emitted from the surface increases due to which current increases. Frequency of incident photons only limits the maximum kinetic energy of the photoelectron emitted and so the current is independent of frequency of photon.

Test: Dual Nature of Radiation & Matter - 1 - Question 27

In various experiments on photo electricity the stopping potential for a given frequency of the incident radiation

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 27

Photoelectric current is zero when the stopping potential is sufficient to repel even the most energetic photoelectrons with the maximum kinetic energy Kmax​ so that Kmax​=eV0​
For a given frequency of the incident radiation, the stopping potential is independent of its intensity.

Test: Dual Nature of Radiation & Matter - 1 - Question 28

In order to increase the kinetic energy of ejected photoelectrons, there should be an increase in

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 28

Relation between kinetic energy and frequency is K.E+ϕ=hν.
Here h is planck's constant, ϕ=constant (Work function)
According to the above equation of photoelectric effect, as we increase frequency of photon, kinetic energy of photoelectron increases. Because the frequency of incident light is directly proportional to kinetic energy.
Hence option C is correct.

Test: Dual Nature of Radiation & Matter - 1 - Question 29

Each photon has the same speed but different

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 29

A photon is a particle of light which essentially is a packet of electromagnetic radiation. The energy of the photon depends on its frequency (how fast the electric field and magnetic field wiggle). The higher the frequency, the more energy the photon has. Of course, a beam of light has many photons. This means that really intense red light (lots of photons, with slightly lower energy) can carry more power to a given area than less intense blue light (fewer photons with higher energy).
The speed of light (c) in a vacuum is constant. This means more energetic (high frequency) photons like X-rays and gamma rays travel at exactly the same speed as lower energy (low frequency) photons, like those in the infrared. As the frequency of a photon goes up, the wavelength goes down, and as the frequency goes down, the wavelength increases.

Test: Dual Nature of Radiation & Matter - 1 - Question 30

Number of ejected photoelectrons increases with increase

Detailed Solution for Test: Dual Nature of Radiation & Matter - 1 - Question 30

The number of electrons ejected can be measured as a function of intensity.
Intensity is equal to energy per unit time per unit area.
Keeping frequency constant, increasing intensity will increase energy thus increase of ejected electrons.
Thus,
Number of ejected photoelectrons will increase with increase in intensity of light.

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