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QUESTION: 1

Work function of a metal is the

Solution:

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. The threshold for this element corresponds to a wavelength of 683 nm.

QUESTION: 2

Photon has a charge of

Solution:

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.

QUESTION: 3

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

Solution:

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

work function=3.3ev

f=3.3×1.6×10^{−19}J /6.626×10^{−34}J−s

=0.8×10^{15}Hz

=8×10^{14}Hz

QUESTION: 4

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

Solution:

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

= 3×10^{8} m/s

and lambda = wavelength

so,

E = [6.63×10^{-34} J. s×3×10^{8} m/s]/5500×10^{-10}m

= [0.36×10^{-26}]/10^{-8}

= 0.36×10^{-18}

= 3.6×10^{-19} J

Or, 2.26ev

QUESTION: 5

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

Solution:

Electromagnetic force on electron will be zero V×B=0

Electrostatic force will be F_{e}=−E backward, hence electron decelerate and velocity will decrease in magnitude.

QUESTION: 6

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

Solution:

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.

QUESTION: 7

Photons can be

Solution:

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.

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

Solution:

Frequency=C/ λ

λ=h/P

frequency=C P/ h

f=(3×10^{8}×3.3×10^{-29})/6.6×10^{-34}

f=3× 10^{13}/2

f=1.5×10^{13} Hz

QUESTION: 9

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

Solution:

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 10^{14} Hz.

QUESTION: 10

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

Solution:

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.

QUESTION: 11

Photoelectric effect is

Solution:

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.

QUESTION: 12

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

Solution:

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.

QUESTION: 13

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

Solution:

Cathode rays are high energetic electron beam.

Specific charge of cathode rays = e/m_{e}

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.

QUESTION: 14

Photoelectric effect is possible

Solution:

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.

QUESTION: 15

Davisson Germer experiment proves that

Solution:

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.

QUESTION: 16

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

Solution:

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

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

Solution:

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

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

Solution:

Given, the maximum kinetic energy: K_{max}=4eV

If V0 be the stopping potential, then K_{max}=eV_{0}

⇒eV_{0}=4eV

⇒V_{0}=4V

QUESTION: 19

In Photoelectric effect

Solution:

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.

QUESTION: 20

A photon is

Solution:

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.

QUESTION: 21

If maximum velocity with which an electron can be emitted from a photo cell is 3.75×10^{8}cms^{−1} then stopping potential is

Solution:

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

Solution:

The maximum kinetic energy for the photoelectrons is

E_{max}=hν−ϕ

where, ν is the frequency of incident light and ϕ is photoelectric work function of metal.

If V_{o} is the stopping potential then

eV_{0}=h(c/λ)−ϕ .....................(since, ν=c/λ)

As per the problem, for incident light of 4000A^{o}, the stopping potential is 2V. When the wavelength of incident light is reduced to 3000A^{o}, then the stopping potential will increase to value more than 2V(as per the above equation).

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

Solution:

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.

QUESTION: 24

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

Solution:

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.

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

Solution:

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

QUESTION: 26

in photoelectric effect, the photoelectric current

Solution:

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.

QUESTION: 27

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

Solution:

Photoelectric current is zero when the stopping potential is sufficient to repel even the most energetic photoelectrons with the maximum kinetic energy K_{max} so that K_{max}=eV_{0}

For a given frequency of the incident radiation, the stopping potential is independent of its intensity.

QUESTION: 28

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

Solution:

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.

QUESTION: 29

Each photon has the same speed but different

Solution:

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.

QUESTION: 30

Number of ejected photoelectrons increases with increase

Solution:

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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