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All questions of Dual Nature of Matter and Radiation for JEE Exam
What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts?a)12.3 nmb)0.127 nmc)123 nmd)1.23 nmCorrect answer is option 'B'. Can you explain this answer?
|
Geetika Shah answered |
and as we know 1Å = 0.1 nm, so 1.227Å = 0.122 nm
Number of ejected photoelectrons increases with increase- a)never
- b)in frequency of light
- c)in wavelength of light
- d)in intensity of light
Correct answer is option 'D'. Can you explain this answer?
Number of ejected photoelectrons increases with increase
a)
never
b)
in frequency of light
c)
in wavelength of light
d)
in intensity of light
|
Rohan Singh answered |
A photon is the smallest possible quantum of light. In general when you turn up the intensity of light you are increasing the number of photons per second that are emitted by the light source. Therefore the intensity of the light can indeed be changed independently of the frequency (or color) of the light.
In order to increase the kinetic energy of ejected photoelectrons, there should be an increase in- a)wavelength of radiation
- b)intensity of radiation
- c)frequency of radiation
- d)Both the wavelength and intensity of radiation
Correct answer is option 'C'. Can you explain this answer?
In order to increase the kinetic energy of ejected photoelectrons, there should be an increase in
a)
wavelength of radiation
b)
intensity of radiation
c)
frequency of radiation
d)
Both the wavelength and intensity of radiation
|
Arjun Singhania answered |
The kinetic energy of emitted photoelectrons should increase with the light amplitude. The rate of electron emission, which is proportional to the measured electric current, should increase as the light frequency is increased.
The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV. What is the stopping potential?- a)30 V
- b)5 V
- c)10 V
- d)15 V
Correct answer is option 'C'. Can you explain this answer?
The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV. What is the stopping potential?
a)
30 V
b)
5 V
c)
10 V
d)
15 V
|
Krishna Iyer answered |
Stopping potential (Vo) is given by
Vo=W/q where W is the work function and q are the charge of an electron.
Given W=20eV−10eV=10eV. Also, q=e
Hence, Vo=(10eV)/e=10V
Vo=W/q where W is the work function and q are the charge of an electron.
Given W=20eV−10eV=10eV. Also, q=e
Hence, Vo=(10eV)/e=10V
Photons of energy 6 eV are incident on a potassium surface of a work function 2.1 eV. What is the stopping potential?- a)-3.9V
- b)-8.1V
- c)-5V
- d)-1.9V
Correct answer is 'A'. Can you explain this answer?
Photons of energy 6 eV are incident on a potassium surface of a work function 2.1 eV. What is the stopping potential?
a)
-3.9V
b)
-8.1V
c)
-5V
d)
-1.9V
|
Priyanka Sharma answered |
From photo-electric equation, eV0= E−φ
eV0=(6−2.1)eV
V0= 3.9 V
stopping potential is a negative potential to stop e- at saturated current .
eV0=(6−2.1)eV
V0= 3.9 V
stopping potential is a negative potential to stop e- at saturated current .
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material .If the frequency is halved and the intensity is doubled, the photoelectric current becomes- a)Doubled
- b)Quadrupled
- c)Halved
- d)zero
Correct answer is option 'D'. Can you explain this answer?
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material .If the frequency is halved and the intensity is doubled, the photoelectric current becomes
a)
Doubled
b)
Quadrupled
c)
Halved
d)
zero
|
Ayush Joshi answered |
If the frequency is halved and intensity is doubled, the frequency of incident light will become 15/2 = 0.75 times the threshold frequency. So, as ν<νo Hence, photoelectric current will be zero.
The work function of a photoelectric material is 3.32 eV. The threshold frequency will be equal to- a)8 ×1014 HZ
- b)9 ×1014 HZ
- c)7 ×1014 HZ
- d)6 ×1014 HZ
Correct answer is option 'A'. Can you explain this answer?
The work function of a photoelectric material is 3.32 eV. The threshold frequency will be equal to
a)
8 ×1014 HZ
b)
9 ×1014 HZ
c)
7 ×1014 HZ
d)
6 ×1014 HZ
|
Anu Sharma answered |
The threshold frequency can be calculated using the formula:
threshold frequency = work function / Planck's constant
Given that the work function is 3.32 eV, we need to convert it to joules by multiplying it by the conversion factor 1.602 x 10^-19 J/eV:
work function = 3.32 eV * 1.602 x 10^-19 J/eV = 5.31264 x 10^-19 J
The value of Planck's constant is 6.626 x 10^-34 J·s.
Therefore, the threshold frequency is:
threshold frequency = 5.31264 x 10^-19 J / (6.626 x 10^-34 J·s)
threshold frequency ≈ 8.03 x 10^14 Hz
So, the threshold frequency will be approximately 8.03 x 10^14 Hz.
threshold frequency = work function / Planck's constant
Given that the work function is 3.32 eV, we need to convert it to joules by multiplying it by the conversion factor 1.602 x 10^-19 J/eV:
work function = 3.32 eV * 1.602 x 10^-19 J/eV = 5.31264 x 10^-19 J
The value of Planck's constant is 6.626 x 10^-34 J·s.
Therefore, the threshold frequency is:
threshold frequency = 5.31264 x 10^-19 J / (6.626 x 10^-34 J·s)
threshold frequency ≈ 8.03 x 10^14 Hz
So, the threshold frequency will be approximately 8.03 x 10^14 Hz.
In a photon-particle collision (such as photon-electron collision) the quantity which is not conserved is- a)total momentum
- b)number of photons
- c)total energy
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
In a photon-particle collision (such as photon-electron collision) the quantity which is not conserved is
a)
total momentum
b)
number of photons
c)
total energy
d)
None of the above
|
|
Rohan Singh answered |
Energy and momentum are conserved, resulting in a reduction of both for the scattered photon. ... This phenomenon could be handled as a collision between two particles—a photon and an electron at rest in the material. Energy and momentum are conserved in the collision.
A photo-sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase the:- a) voltage applied to the light source
- b) intensity of light
- c) wavelength of light
- d) frequency of light
Correct answer is option 'D'. Can you explain this answer?
A photo-sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase the:
a)
voltage applied to the light source
b)
intensity of light
c)
wavelength of light
d)
frequency of light
|
|
Rohan Singh answered |
The emission of photoelectron takes place only, when the frequency of the incident light is above a certain critical value, characteristic of that metal. The critical value of frequency is known as the threshold frequency for the metal of the emitting electrode.
Suppose that when light of certain frequency is incident over a metal surface, the photo- electrons are emitted. To take photoelectric current zero, a particular value of stopping potential will be needed. If we go on reducing the frequency of incident light, the value of stopping potential will also go on decreasing. At certain value of frequency v0, the photoelectric current will become zero, even when no retarding potential is applied. This frequency vq corresponds to the threshold for the metal surface. The emission of photoelectrons does not take place, till frequency of incident light is below this value.
Can you explain the answer of this question below:Light from a bulb is falling on a wooden table but no photo electrons are emitted as
- A:
Work function of wood is less
- B:
Work function of wood is more
- C:
It depends on the frequency
- D:
It is independent of work function
The answer is b.
Light from a bulb is falling on a wooden table but no photo electrons are emitted as
Work function of wood is less
Work function of wood is more
It depends on the frequency
It is independent of work function
|
Rishika Patel answered |
Explanation:
When light falls on a metal surface, electrons may be emitted from the metal surface. This phenomenon is known as the photoelectric effect. The electrons emitted from the metal surface are called photoelectrons.
The photoelectric effect can be explained by considering that light is made up of photons. Each photon has a certain amount of energy, given by its frequency. When a photon strikes a metal surface, its energy can be transferred to an electron in the metal. If the energy of the photon is greater than the work function of the metal, the electron can be emitted from the metal surface.
In the case of a wooden table, the work function of the wood is more than the energy of the photons of the light falling on it. Therefore, no photoelectrons are emitted. This is because the energy of the photons is not enough to overcome the work function of the wood.
Key Points:
- The photoelectric effect is the emission of electrons from a metal surface when light falls on it.
- The energy of a photon is given by its frequency.
- If the energy of a photon is greater than the work function of the metal, electrons can be emitted from the metal surface.
- In the case of a wooden table, the work function of the wood is more than the energy of the photons of the light falling on it, so no photoelectrons are emitted.
When light falls on a metal surface, electrons may be emitted from the metal surface. This phenomenon is known as the photoelectric effect. The electrons emitted from the metal surface are called photoelectrons.
The photoelectric effect can be explained by considering that light is made up of photons. Each photon has a certain amount of energy, given by its frequency. When a photon strikes a metal surface, its energy can be transferred to an electron in the metal. If the energy of the photon is greater than the work function of the metal, the electron can be emitted from the metal surface.
In the case of a wooden table, the work function of the wood is more than the energy of the photons of the light falling on it. Therefore, no photoelectrons are emitted. This is because the energy of the photons is not enough to overcome the work function of the wood.
Key Points:
- The photoelectric effect is the emission of electrons from a metal surface when light falls on it.
- The energy of a photon is given by its frequency.
- If the energy of a photon is greater than the work function of the metal, electrons can be emitted from the metal surface.
- In the case of a wooden table, the work function of the wood is more than the energy of the photons of the light falling on it, so no photoelectrons are emitted.
Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential is- a)Greater with visible light
- b)Equal in both cases
- c)Greater with ultraviolet light
- d)Infinite in both cases
Correct answer is 'C'. Can you explain this answer?
Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential is
a)
Greater with visible light
b)
Equal in both cases
c)
Greater with ultraviolet light
d)
Infinite in both cases
|
|
Riya Banerjee answered |
λ for U.V is less than λ for visible light
ν for U.V is greater than ν for visible light
∴ potential is greater for U.V light.
as K.Eα1/λ
ν for U.V is greater than ν for visible light
∴ potential is greater for U.V light.
as K.Eα1/λ
An α – particle and a deutron are accelerated through the same potential difference. What will be the ratio of their de-Broglie wavelength?- a)1 / 3
- b)1 / 5
- c)1 / 4
- d)1 / 2
Correct answer is 'D'. Can you explain this answer?
An α – particle and a deutron are accelerated through the same potential difference. What will be the ratio of their de-Broglie wavelength?
a)
1 / 3
b)
1 / 5
c)
1 / 4
d)
1 / 2
|
|
Abhay Iyer answered |
Mass of alpha(a)= 4.
mass of deutron(d) =2.
wavelength=h/mv
Potential are same of both particle so their speed will be same .
Here, wavelength is inversely proportional to mass
so,
wavelength of alpha/wavelength of deutron = mass of deutron /mass of alpha.= 2/4=1/2
mass of deutron(d) =2.
wavelength=h/mv
Potential are same of both particle so their speed will be same .
Here, wavelength is inversely proportional to mass
so,
wavelength of alpha/wavelength of deutron = mass of deutron /mass of alpha.= 2/4=1/2
The de-Broglie wavelength of an electron is 1.0 nm. What is the retarding potential required to stop it?- a)1.5 V
- b)3.5 V
- c)6.5 V
- d)4.5 V
Correct answer is option 'A'. Can you explain this answer?
The de-Broglie wavelength of an electron is 1.0 nm. What is the retarding potential required to stop it?
a)
1.5 V
b)
3.5 V
c)
6.5 V
d)
4.5 V
|
Neha Sharma answered |
λ=h/P
Here P= √2mk
K=kinetic energy
λ=h/√2mk
By energy conversion
K=eVs
λ=h/√2meVs
√Vs=h/ λ√2me
Vs=h2/ λ2me
Vs=(6.626x10-34)2/(1x10-9)2x2x9.11x10-31xe
Where e=1.602x10-19
Vs=43.9x10-68/29.1x10-68
=1.508v
Vs=1.5v
Here P= √2mk
K=kinetic energy
λ=h/√2mk
By energy conversion
K=eVs
λ=h/√2meVs
√Vs=h/ λ√2me
Vs=h2/ λ2me
Vs=(6.626x10-34)2/(1x10-9)2x2x9.11x10-31xe
Where e=1.602x10-19
Vs=43.9x10-68/29.1x10-68
=1.508v
Vs=1.5v
In an experiment of photoelectric emission for incident light of 4000 A0, the stopping potential is 2V. If the wavelength of incident light is made 3000 A0, then stopping potential will be- a)zero
- b)more than 2 volt
- c)2 Volt
- d)less than 2 volt
Correct answer is option 'B'. Can you explain this answer?
In an experiment of photoelectric emission for incident light of 4000 A0, the stopping potential is 2V. If the wavelength of incident light is made 3000 A0, then stopping potential will be
a)
zero
b)
more than 2 volt
c)
2 Volt
d)
less than 2 volt
|
|
Geetika Shah answered |
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).
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).
Light of wavelength 4000 Â is incident on a metal of work function 3.2 x 10-19 J. What is the maximum kinetic energy of the emitted electron?
- a)2.2 x 10-19 J
- b)1.75 x 10-19 J
- c)0.75 x 10-19 J
- d)1.1 x 10-19 J
Correct answer is option 'B'. Can you explain this answer?
Light of wavelength 4000 Â is incident on a metal of work function 3.2 x 10-19 J. What is the maximum kinetic energy of the emitted electron?
a)
2.2 x 10-19 J
b)
1.75 x 10-19 J
c)
0.75 x 10-19 J
d)
1.1 x 10-19 J
|
Om Desai answered |
Given,
λ= 4000 Å ;W=3.2× 10-19 j
as we know that
K.E = (hc/ λ) - W
= (6.6× 10-34 × 3 × 108/ 4000×10-10 )- 3.20× 10-19
= 1.75 × 10-19
λ= 4000 Å ;W=3.2× 10-19 j
as we know that
K.E = (hc/ λ) - W
= (6.6× 10-34 × 3 × 108/ 4000×10-10 )- 3.20× 10-19
= 1.75 × 10-19
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- a)1.64 eV
- b)1.56 eV
- c)1.52 eV
- d)1.41 eV
Correct answer is option 'A'. Can you explain this answer?
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
a)
1.64 eV
b)
1.56 eV
c)
1.52 eV
d)
1.41 eV
|
|
Neha Sharma answered |
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
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
In which case is electron emission from a metal not known?- a)Illuminating a metal surface with suitable light
- b)Applying a very strong magnetic field to a metal
- c)Applying a very strong electric field to a metal
- d)Heating the metal sufficiently
Correct answer is option 'B'. Can you explain this answer?
In which case is electron emission from a metal not known?
a)
Illuminating a metal surface with suitable light
b)
Applying a very strong magnetic field to a metal
c)
Applying a very strong electric field to a metal
d)
Heating the metal sufficiently
|
Divey Sethi answered |
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.
Applying a very strong magnetic field to a metal we don’t know the electron emission.
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- a)1.6 ×1013Hz
- b)1.5 ×1013Hz
- c)1.7 ×1013Hz
- d)1.8 ×1013Hz
Correct answer is option 'B'. Can you explain this answer?
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
a)
1.6 ×1013Hz
b)
1.5 ×1013Hz
c)
1.7 ×1013Hz
d)
1.8 ×1013Hz
|
|
Lavanya Menon answered |
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
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
Of the following moving with the same momentum, the one which has the largest wavelength is- a)a proton
- b)α -particle
- c)an electron
- d)all have the same de Broglie wavelength
Correct answer is option 'D'. Can you explain this answer?
Of the following moving with the same momentum, the one which has the largest wavelength is
a)
a proton
b)
α -particle
c)
an electron
d)
all have the same de Broglie wavelength
|
|
Gaurav Kumar answered |
De Broglie wavelength = h/mv,
where h = plancks constant, mv = momentum.
As they move with same momentum, de Broglie wavelength remains constant for all.
where h = plancks constant, mv = momentum.
As they move with same momentum, de Broglie wavelength remains constant for all.
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- a)2 V
- b)6 V
- c)4 V
- d)10 V
Correct answer is option 'C'. Can you explain this answer?
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
a)
2 V
b)
6 V
c)
4 V
d)
10 V
|
|
Rajeev Saxena answered |
Stopping potential is nothing but the maximum kinetic energy of electros which get emitted during photoelectric effect.
So the stopping potential = maximum kinetic energy = 4eV
In various experiments on photo electricity the stopping potential for a given frequency of the incident radiation- a)is independent of the radiation intensity
- b)is inversely proportional to radiation intensity
- c)is proportional to radiation intensity
- d)none
Correct answer is option 'A'. Can you explain this answer?
In various experiments on photo electricity the stopping potential for a given frequency of the incident radiation
a)
is independent of the radiation intensity
b)
is inversely proportional to radiation intensity
c)
is proportional to radiation intensity
d)
none
|
|
Shreya Singh answered |
If the voltage across the electrodes of a cathode ray tube is 500 volts then energy gained by the electrons is- a)7 ×10-17 J
- b)8 ×10−17 J
- c)6 ×10−17 J
- d)9 ×10−17 J
Correct answer is option 'B'. Can you explain this answer?
If the voltage across the electrodes of a cathode ray tube is 500 volts then energy gained by the electrons is
a)
7 ×10-17 J
b)
8 ×10−17 J
c)
6 ×10−17 J
d)
9 ×10−17 J
|
|
Om Desai answered |
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
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
Photons can be- a)scattered
- b)deflected by magnetic fields
- c)deflected by electric fields
- d)none
Correct answer is option 'A'. Can you explain this answer?
Photons can be
a)
scattered
b)
deflected by magnetic fields
c)
deflected by electric fields
d)
none
|
|
Rajeev Saxena answered |
As this electron changes orbit, its energy is reduced, and the excess energy is given off in the form of a photon, called a “characteristic photon.” In pair production, photon energies greater than 1.02 MeV interact with the strong electric field of the nucleus and lose all incident energy.
Light of two different frequencies whose photons have energies of 1eV and 2.5 eV respectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons is- a)1:5
- b)1:4
- c)1:3
- d)1:2
Correct answer is option 'D'. Can you explain this answer?
Light of two different frequencies whose photons have energies of 1eV and 2.5 eV respectively successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of emitted electrons is
a)
1:5
b)
1:4
c)
1:3
d)
1:2
|
|
Shreya Gupta answered |
In Photoelectric effect- a)electrical energy is converted magnetic field energy
- b)electrical energy is converted into light energy
- c)light is converted into electrical energy
- d)electrical energy is converted into heat
Correct answer is option 'C'. Can you explain this answer?
In Photoelectric effect
a)
electrical energy is converted magnetic field energy
b)
electrical energy is converted into light energy
c)
light is converted into electrical energy
d)
electrical energy is converted into heat
|
|
Nikita Singh answered |
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.
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.
A photon is- a)a positive charged particle
- b)a quantum of light energy
- c)an instrument for measuring light intensity
- d)a quantum of matter
Correct answer is option 'B'. Can you explain this answer?
A photon is
a)
a positive charged particle
b)
a quantum of light energy
c)
an instrument for measuring light intensity
d)
a quantum of matter
|
|
Rajat Patel answered |
The photon is a type of elementary particle, the quantum of the electromagnetic field including electromagnetic radiation such as light, and the force carrier for the electromagnetic force (even when static via virtual particles). ... The quanta in a light wave are not spatially localized.
in photoelectric effect, the photoelectric current- a)both on intensity and frequency of incident beam
- b)increases when frequency of incident photons increases
- c)decreases when frequency of incident photons increases
- d)does not depend on photon frequency but only on intensity of incident
Correct answer is option 'D'. Can you explain this answer?
in photoelectric effect, the photoelectric current
a)
both on intensity and frequency of incident beam
b)
increases when frequency of incident photons increases
c)
decreases when frequency of incident photons increases
d)
does not depend on photon frequency but only on intensity of incident
|
|
Arun Khanna answered |
In photoelectric effect,when light incident on metal surface the no. Of electron emitted is depends on intensity of light and speed of Electron is depends on frequency of incidents light. Photoelectric current is due to the ejection of photo electrons.
Photoelectric effect is- a)emission of electrons by glass when illuminated by light of suitable frequency
- b)emission of electrons by metals when heated
- c)emission of electrons by insulators when illuminated by light of suitable frequency
- d)emission of electrons by metals when illuminated by light of suitable frequency
Correct answer is option 'D'. Can you explain this answer?
Photoelectric effect is
a)
emission of electrons by glass when illuminated by light of suitable frequency
b)
emission of electrons by metals when heated
c)
emission of electrons by insulators when illuminated by light of suitable frequency
d)
emission of electrons by metals when illuminated by light of suitable frequency
|
Prasenjit Malik answered |
Understanding the Photoelectric Effect
The photoelectric effect is a fundamental phenomenon in physics, describing how light interacts with matter, particularly metals. Here’s a detailed explanation of why option 'D' is the correct choice.
Definition of the Photoelectric Effect
- The photoelectric effect refers to the emission of electrons from a metal surface when it is exposed to light of suitable frequency (or energy).
Key Characteristics
- Electron Emission: When light shines on a metal, photons (light particles) can impart energy to the electrons in the metal.
- Threshold Frequency: Each metal has a specific threshold frequency. If the frequency of the incoming light is higher than this threshold, electrons are emitted. If it is lower, no emission occurs regardless of light intensity.
Why Option 'D' is Correct
- Metals and Electron Emission: Metals have free electrons that can be easily displaced. When illuminated by light of suitable frequency, these electrons can absorb sufficient energy to overcome the binding forces holding them in the metal, leading to their emission.
- Other Options Explained:
- Option A: Glass does not emit electrons; it is an insulator.
- Option B: Heating metals causes thermal emission, not the photoelectric effect.
- Option C: Insulators do not typically emit electrons upon illumination because their electrons are tightly bound.
Conclusion
- The photoelectric effect is a crucial concept in understanding quantum mechanics and the dual nature of light. It highlights how light can behave as both a wave and a particle, contributing to advancements in technology, such as photoelectric cells and solar panels. Thus, option 'D' accurately describes this phenomenon.
The photoelectric effect is a fundamental phenomenon in physics, describing how light interacts with matter, particularly metals. Here’s a detailed explanation of why option 'D' is the correct choice.
Definition of the Photoelectric Effect
- The photoelectric effect refers to the emission of electrons from a metal surface when it is exposed to light of suitable frequency (or energy).
Key Characteristics
- Electron Emission: When light shines on a metal, photons (light particles) can impart energy to the electrons in the metal.
- Threshold Frequency: Each metal has a specific threshold frequency. If the frequency of the incoming light is higher than this threshold, electrons are emitted. If it is lower, no emission occurs regardless of light intensity.
Why Option 'D' is Correct
- Metals and Electron Emission: Metals have free electrons that can be easily displaced. When illuminated by light of suitable frequency, these electrons can absorb sufficient energy to overcome the binding forces holding them in the metal, leading to their emission.
- Other Options Explained:
- Option A: Glass does not emit electrons; it is an insulator.
- Option B: Heating metals causes thermal emission, not the photoelectric effect.
- Option C: Insulators do not typically emit electrons upon illumination because their electrons are tightly bound.
Conclusion
- The photoelectric effect is a crucial concept in understanding quantum mechanics and the dual nature of light. It highlights how light can behave as both a wave and a particle, contributing to advancements in technology, such as photoelectric cells and solar panels. Thus, option 'D' accurately describes this phenomenon.
If the work function of a material is 2eV, then minimum frequency of light required to emit photo-electrons is - a)5.0 ×1015Hz
- b)4.6 ×1015Hz
- c)4.6 ×1014Hz
- d)5.0 ×1014Hz
Correct answer is option 'C'. Can you explain this answer?
If the work function of a material is 2eV, then minimum frequency of light required to emit photo-electrons is
a)
5.0 ×1015Hz
b)
4.6 ×1015Hz
c)
4.6 ×1014Hz
d)
5.0 ×1014Hz
|
|
Lavanya Menon answered |
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.
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
- a)photocurrent decreases nonlinearly with intensity
- b)photocurrent decreases linearly with intensity
- c)photocurrent remains same with intensity
- d)photocurrent increases linearly with intensity
Correct answer is option 'D'. Can you explain this answer?
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
a)
photocurrent decreases nonlinearly with intensity
b)
photocurrent decreases linearly with intensity
c)
photocurrent remains same with intensity
d)
photocurrent increases linearly with intensity
|
|
Nikita Singh answered |
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.
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.
Energy of a photon of green light of wavelength 5500 is (given: h = 6.62 ×10−34Js−1) approximately- a)3.01eV
- b)2.81ev
- c)2.26 eV
- d)2.93 eV
Correct answer is 'C'. Can you explain this answer?
Energy of a photon of green light of wavelength 5500 is (given: h = 6.62 ×10−34Js−1) approximately
a)
3.01eV
b)
2.81ev
c)
2.26 eV
d)
2.93 eV
|
|
Rajesh Gupta answered |
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
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
The photoelectric threshold frequency of a metal is f0. When light of frequency 4f0is incident on the metal, the maximum K.E. of the emitted electron is- a)2 hf0
- b)4 hf0
- c)3 hf0
- d)hf0/ 2
Correct answer is option 'C'. Can you explain this answer?
The photoelectric threshold frequency of a metal is f0. When light of frequency 4f0is incident on the metal, the maximum K.E. of the emitted electron is
a)
2 hf0
b)
4 hf0
c)
3 hf0
d)
hf0/ 2
|
Hansa Sharma answered |
The maximum kinetic energy of the emitted electrons is given by
Kmax=hυ−ϕ0=h(4υ)−h(υ)=3hυ
Kmax=hυ−ϕ0=h(4υ)−h(υ)=3hυ
The work function of metal is 1 eV. Light of wavelength 3000 Å is incident on this metal surface. The velocity of emitted photo-electrons will be- a) 10 m/sec
- b)1 × 103 m / sec
- c)1 × 104 m / sec
- d)1× 106 m / sec
Correct answer is option 'D'. Can you explain this answer?
The work function of metal is 1 eV. Light of wavelength 3000 Å is incident on this metal surface. The velocity of emitted photo-electrons will be
a)
10 m/sec
b)
1 × 103 m / sec
c)
1 × 104 m / sec
d)
1× 106 m / sec
|
|
Sai Mehta answered |
Understanding the Problem
The work function of the metal is given as 1 eV, and light of wavelength 3000 Å (Angstroms) is incident on it. We need to find the velocity of the emitted photoelectrons.
Key Concepts
- Work Function (Φ): The minimum energy required to remove an electron from the surface of a metal, given in electron volts (eV).
- Energy of Photons (E): The energy of the incident light can be calculated using the formula:
E = hc/λ
where:
- h = Planck's constant (4.135667696 × 10^-15 eV·s)
- c = Speed of light (3 × 10^8 m/s)
- λ = Wavelength (in meters)
Calculating Energy of Incident Light
1. Convert Wavelength:
3000 Å = 3000 × 10^-10 m = 3 × 10^-7 m
2. Calculate Energy (E):
E = (4.135667696 × 10^-15 eV·s) × (3 × 10^8 m/s) / (3 × 10^-7 m)
E = 4.135667696 eV
Photoelectric Effect
- Kinetic Energy (KE): The kinetic energy of the emitted electrons is given by:
KE = E - Φ
KE = 4.135667696 eV - 1 eV = 3.135667696 eV
Calculating Velocity of Electrons
- Using Kinetic Energy Formula:
KE = (1/2)mv², where m is the mass of an electron (approximately 9.11 × 10^-31 kg).
- Rearranging gives:
v = sqrt(2 * KE / m)
- Converting KE to Joules (1 eV = 1.6 × 10^-19 J):
KE = 3.135667696 eV × 1.6 × 10^-19 J/eV = 5.016 e-19 J
- Now, substituting the values:
v = sqrt(2 * 5.016 e-19 J / 9.11 e-31 kg)
v ≈ 1 × 10^6 m/s
Conclusion
Thus, the velocity of the emitted photoelectrons is approximately 1 × 10^6 m/s, confirming that the correct answer is option 'D'.
The work function of the metal is given as 1 eV, and light of wavelength 3000 Å (Angstroms) is incident on it. We need to find the velocity of the emitted photoelectrons.
Key Concepts
- Work Function (Φ): The minimum energy required to remove an electron from the surface of a metal, given in electron volts (eV).
- Energy of Photons (E): The energy of the incident light can be calculated using the formula:
E = hc/λ
where:
- h = Planck's constant (4.135667696 × 10^-15 eV·s)
- c = Speed of light (3 × 10^8 m/s)
- λ = Wavelength (in meters)
Calculating Energy of Incident Light
1. Convert Wavelength:
3000 Å = 3000 × 10^-10 m = 3 × 10^-7 m
2. Calculate Energy (E):
E = (4.135667696 × 10^-15 eV·s) × (3 × 10^8 m/s) / (3 × 10^-7 m)
E = 4.135667696 eV
Photoelectric Effect
- Kinetic Energy (KE): The kinetic energy of the emitted electrons is given by:
KE = E - Φ
KE = 4.135667696 eV - 1 eV = 3.135667696 eV
Calculating Velocity of Electrons
- Using Kinetic Energy Formula:
KE = (1/2)mv², where m is the mass of an electron (approximately 9.11 × 10^-31 kg).
- Rearranging gives:
v = sqrt(2 * KE / m)
- Converting KE to Joules (1 eV = 1.6 × 10^-19 J):
KE = 3.135667696 eV × 1.6 × 10^-19 J/eV = 5.016 e-19 J
- Now, substituting the values:
v = sqrt(2 * 5.016 e-19 J / 9.11 e-31 kg)
v ≈ 1 × 10^6 m/s
Conclusion
Thus, the velocity of the emitted photoelectrons is approximately 1 × 10^6 m/s, confirming that the correct answer is option 'D'.
An X-ray tube is operating at 50 kV and 20 mA. The target material of the tube has a mass of 1.0 kg and specific heat 495 J kg–1°C–1. One percent of the supplied electric power is converted into X-rays and the entire remaining energy goes into heating the target. Then- a)A suitable target material must have a high melting temperature
- b)A suitable target material must have low thermal conductivity
- c)The average rate of rise of temperature of target would be 2 °C/s
- d)The minimum wavelength of the X-rays emitted is about 0.25 x 10-10 m
Correct answer is option 'A,C,D'. Can you explain this answer?
An X-ray tube is operating at 50 kV and 20 mA. The target material of the tube has a mass of 1.0 kg and specific heat 495 J kg–1°C–1. One percent of the supplied electric power is converted into X-rays and the entire remaining energy goes into heating the target. Then
a)
A suitable target material must have a high melting temperature
b)
A suitable target material must have low thermal conductivity
c)
The average rate of rise of temperature of target would be 2 °C/s
d)
The minimum wavelength of the X-rays emitted is about 0.25 x 10-10 m
|
Arya Chakraborty answered |
Operating Conditions of the X-ray Tube
The X-ray tube operates at specific conditions that influence the choice of target material and its thermal behavior:
- Voltage and Current: The tube operates at 50 kV and 20 mA, which determines the power output.
- Power Conversion: Only 1% of the supplied electric power is converted into X-rays, while the rest contributes to heating the target.
Target Material Requirements
1. High Melting Temperature:
- A target material must have a high melting point to withstand the intense heat generated during operation.
- Materials like tungsten, commonly used for X-ray targets, have high melting temperatures (~3422°C), making them suitable for this application.
2. Low Thermal Conductivity:
- This statement is incorrect. A suitable target material should have high thermal conductivity to dissipate heat efficiently, preventing overheating and maintaining performance.
Rate of Temperature Rise
- Temperature Calculation:
- The electric power supplied to the tube is calculated as:
Power (P) = Voltage (V) × Current (I) = 50,000 V × 0.02 A = 1000 W.
- With 99% going into heating, the heating power = 0.99 × 1000 W = 990 W.
- The rate of temperature rise (dT/dt) can be calculated using the formula:
dT/dt = Power / (mass × specific heat) = 990 W / (1.0 kg × 495 J/kg°C) = 2 °C/s.
Minimum Wavelength of X-rays
- Calculation of Minimum Wavelength:
- The minimum wavelength (λ) of the emitted X-rays can be determined using the formula:
λ = hc/eV, where h is Planck's constant, c is the speed of light, and eV is the energy in electron volts.
- Substituting the values for 50 kV gives approximately λ = 0.25 × 10^-10 m.
Conclusion
- The correct options are A, C, and D, highlighting the necessary properties of target materials and the operational characteristics of the X-ray tube.
The X-ray tube operates at specific conditions that influence the choice of target material and its thermal behavior:
- Voltage and Current: The tube operates at 50 kV and 20 mA, which determines the power output.
- Power Conversion: Only 1% of the supplied electric power is converted into X-rays, while the rest contributes to heating the target.
Target Material Requirements
1. High Melting Temperature:
- A target material must have a high melting point to withstand the intense heat generated during operation.
- Materials like tungsten, commonly used for X-ray targets, have high melting temperatures (~3422°C), making them suitable for this application.
2. Low Thermal Conductivity:
- This statement is incorrect. A suitable target material should have high thermal conductivity to dissipate heat efficiently, preventing overheating and maintaining performance.
Rate of Temperature Rise
- Temperature Calculation:
- The electric power supplied to the tube is calculated as:
Power (P) = Voltage (V) × Current (I) = 50,000 V × 0.02 A = 1000 W.
- With 99% going into heating, the heating power = 0.99 × 1000 W = 990 W.
- The rate of temperature rise (dT/dt) can be calculated using the formula:
dT/dt = Power / (mass × specific heat) = 990 W / (1.0 kg × 495 J/kg°C) = 2 °C/s.
Minimum Wavelength of X-rays
- Calculation of Minimum Wavelength:
- The minimum wavelength (λ) of the emitted X-rays can be determined using the formula:
λ = hc/eV, where h is Planck's constant, c is the speed of light, and eV is the energy in electron volts.
- Substituting the values for 50 kV gives approximately λ = 0.25 × 10^-10 m.
Conclusion
- The correct options are A, C, and D, highlighting the necessary properties of target materials and the operational characteristics of the X-ray tube.
The specific charge for positive rays is much less than that for cathode rays. This is because- a)experimental approach is defective
- b)their masses are very large
- c)their charge is very small
- d)they are positively charged
Correct answer is option 'B'. Can you explain this answer?
The specific charge for positive rays is much less than that for cathode rays. This is because
a)
experimental approach is defective
b)
their masses are very large
c)
their charge is very small
d)
they are positively charged
|
Vijay Bansal answered |
Specific charge = charge/Mass The positive rays are stream of positive ions. The mass of positive ion is much more than that of electrons, hence specific charge of positive ions is less
Uniform electric and magnetic fields are produced pointing in the same direction. An electron is projected pointing in the same direction, then- a)the electron turns to right
- b)the velocity of electron increases
- c)the electron turns to left
- d)the velocity of electron decreases
Correct answer is option 'D'. Can you explain this answer?
Uniform electric and magnetic fields are produced pointing in the same direction. An electron is projected pointing in the same direction, then
a)
the electron turns to right
b)
the velocity of electron increases
c)
the electron turns to left
d)
the velocity of electron decreases
|
|
Divey Sethi answered |
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.
Electrostatic force will be Fe=−E backward, hence electron decelerate and velocity will decrease in magnitude.
Each photon has the same speed but different- a)frequency
- b)radius
- c)rest mass
- d)energy
Correct answer is option 'A'. Can you explain this answer?
Each photon has the same speed but different
a)
frequency
b)
radius
c)
rest mass
d)
energy
|
|
Arjun Singhania answered |
The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.
Work function of a metal is the
- a)maximum energy required by an electron to escape from the metal surface
- b)minimum energy required by an electron to escape from the metal surface
- c)energy required by an electron to escape from the metal surface
- d)energy required by an electron to get absorbed in the metal surface
Correct answer is option 'B'. Can you explain this answer?
Work function of a metal is the
a)
maximum energy required by an electron to escape from the metal surface
b)
minimum energy required by an electron to escape from the metal surface
c)
energy required by an electron to escape from the metal surface
d)
energy required by an electron to get absorbed in the metal surface
|
Madhavan Patel answered |
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.
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.
Photon has a charge of- a)-2e; e= charge on electron
- b)+2e; e= charge on electron
- c)zero
- d)-e; e= charge on electron
Correct answer is option 'C'. Can you explain this answer?
Photon has a charge of
a)
-2e; e= charge on electron
b)
+2e; e= charge on electron
c)
zero
d)
-e; e= charge on electron
|
|
Alok Mehta answered |
A photon is massless, has no electric charge, and is a stable particle. A photon has two possible polarization states. In the momentum representation of the photon, which is preferred in quantum field theory, a photon is described by its wave vector, which determines its wavelength λ and its direction of propagation.
A particle of mass M at rest decays into two particles of masses m1 and m2, having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, l1/l2 is- a)m1/m2
- b)m2/m1
- c)1.0
- d)√m2/m1
Correct answer is option 'C'. Can you explain this answer?
A particle of mass M at rest decays into two particles of masses m1 and m2, having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, l1/l2 is
a)
m1/m2
b)
m2/m1
c)
1.0
d)
√m2/m1
|
Bs Academy answered |
By law of conservation of momentum
–ve sign indicates that both the particles are moving in opposite direction. Now de-Broglie wavelengths
–ve sign indicates that both the particles are moving in opposite direction. Now de-Broglie wavelengths
In a photoemissive cell with executing wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be- a)v(3/4)1/2
- b)v(4/3)1/2
- c)Less than v(4/3)1/2
- d)Greater than v(4/3)1/2
Correct answer is option 'D'. Can you explain this answer?
In a photoemissive cell with executing wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be
a)
v(3/4)1/2
b)
v(4/3)1/2
c)
Less than v(4/3)1/2
d)
Greater than v(4/3)1/2
|
Infinity Academy answered |
A proton, a deuteron and an a-particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is
- a)1 : 2 : 8
- b)1 : 2 : 4
- c)1 : 1 : 2
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
A proton, a deuteron and an a-particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is
a)
1 : 2 : 8
b)
1 : 2 : 4
c)
1 : 1 : 2
d)
None of these
|
Top Rankers answered |
The deflection suffered by charged particle in an electric field is
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