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JEE Main Previous Year Questions (2025): Dual Nature of Matter and Radiation

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JEE Main Previous Year Questions 
(2025): Dual Nature of Matter and 
Radiation 
Q1: An electron in the ground state of the hydrogen atom has the orbital radius of 
?? .?? × ????
-????
 ?? while that for the electron in third excited state is ?? .???? × ????
-????
 ?? . 
The ratio of the de Broglie wavelengths of electron in the ground state to that in the 
excited state is 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 4 
B. 3 
C. 9 
D. 16 
Ans: A 
Solution: 
We know, ?? =
h
????
 
and ?????? =
?? h
2?? 
? ???? =
?? h
2????
 
So, ?? =
h
?? h
2???? = 2?? ?? ?? 
? ?? ?
?? ?? 
We can write, 
?? ?? ?? ?? = (
?? ?? ?? ?? )(
?? ?? ?? ?? ) 
?
?? ?? ?? ?? = (
5.3× 10
-11
84.8× 10
-11
)(
4
1
) =
1
4
 
?
?? ?? ?? ?? = 4 
Q2: The work functions of cesium ( Cs ) and lithium ( Li ) metals are 1.9 eV and 2.5 eV , 
respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, 
then photo-electric effect is possible for the case of 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Both Cs and Li 
B. Neither Cs nor Li 
C. Li only 
D. Cs only 
Page 2


JEE Main Previous Year Questions 
(2025): Dual Nature of Matter and 
Radiation 
Q1: An electron in the ground state of the hydrogen atom has the orbital radius of 
?? .?? × ????
-????
 ?? while that for the electron in third excited state is ?? .???? × ????
-????
 ?? . 
The ratio of the de Broglie wavelengths of electron in the ground state to that in the 
excited state is 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 4 
B. 3 
C. 9 
D. 16 
Ans: A 
Solution: 
We know, ?? =
h
????
 
and ?????? =
?? h
2?? 
? ???? =
?? h
2????
 
So, ?? =
h
?? h
2???? = 2?? ?? ?? 
? ?? ?
?? ?? 
We can write, 
?? ?? ?? ?? = (
?? ?? ?? ?? )(
?? ?? ?? ?? ) 
?
?? ?? ?? ?? = (
5.3× 10
-11
84.8× 10
-11
)(
4
1
) =
1
4
 
?
?? ?? ?? ?? = 4 
Q2: The work functions of cesium ( Cs ) and lithium ( Li ) metals are 1.9 eV and 2.5 eV , 
respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, 
then photo-electric effect is possible for the case of 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Both Cs and Li 
B. Neither Cs nor Li 
C. Li only 
D. Cs only 
Ans: D 
Solution: 
For photo-electric effect, energy of the photon must be greater than the work function of the 
metal. 
We know, ?? =
1240
?? =
1240
550
 (as ?? = 550 nm given) 
? ?? = 2.25eV 
here, ?? > ?? ????
 
So, ?? ?? only. 
Q3: A light source of wavelength ?? illuminates a metal surface and electrons are 
ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a 
light source of wavelength 
?? ?? , then the maximum kinetic energy of ejected electrons 
will be (The work function of metal is ?? ???? ) 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 5 eV 
B. 3 eV 
C. 2 eV 
D. 6 eV 
Ans: A 
Solution: 
Let's break down the problem step by step. 
The photoelectric equation is given by: 
?? max
=
h?? ?? - ?? 
where: 
?? max
 is the maximum kinetic energy of the electrons. 
h?? ?? is the energy of the incident photon. 
?? is the work function of the metal. 
For the initial light source of wavelength ?? , we know: 
2eV=
h?? ?? - 1eV 
Solving for 
h?? ?? gives: 
h?? ?? = 2eV+ 1eV= 3eV 
Now, if the wavelength is halved to 
?? 2
, the photon energy becomes: 
h?? ?? /2
=
2h?? ?? = 2× 3eV= 6eV 
The maximum kinetic energy for the new wavelength is then: 
Page 3


JEE Main Previous Year Questions 
(2025): Dual Nature of Matter and 
Radiation 
Q1: An electron in the ground state of the hydrogen atom has the orbital radius of 
?? .?? × ????
-????
 ?? while that for the electron in third excited state is ?? .???? × ????
-????
 ?? . 
The ratio of the de Broglie wavelengths of electron in the ground state to that in the 
excited state is 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 4 
B. 3 
C. 9 
D. 16 
Ans: A 
Solution: 
We know, ?? =
h
????
 
and ?????? =
?? h
2?? 
? ???? =
?? h
2????
 
So, ?? =
h
?? h
2???? = 2?? ?? ?? 
? ?? ?
?? ?? 
We can write, 
?? ?? ?? ?? = (
?? ?? ?? ?? )(
?? ?? ?? ?? ) 
?
?? ?? ?? ?? = (
5.3× 10
-11
84.8× 10
-11
)(
4
1
) =
1
4
 
?
?? ?? ?? ?? = 4 
Q2: The work functions of cesium ( Cs ) and lithium ( Li ) metals are 1.9 eV and 2.5 eV , 
respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, 
then photo-electric effect is possible for the case of 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Both Cs and Li 
B. Neither Cs nor Li 
C. Li only 
D. Cs only 
Ans: D 
Solution: 
For photo-electric effect, energy of the photon must be greater than the work function of the 
metal. 
We know, ?? =
1240
?? =
1240
550
 (as ?? = 550 nm given) 
? ?? = 2.25eV 
here, ?? > ?? ????
 
So, ?? ?? only. 
Q3: A light source of wavelength ?? illuminates a metal surface and electrons are 
ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a 
light source of wavelength 
?? ?? , then the maximum kinetic energy of ejected electrons 
will be (The work function of metal is ?? ???? ) 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 5 eV 
B. 3 eV 
C. 2 eV 
D. 6 eV 
Ans: A 
Solution: 
Let's break down the problem step by step. 
The photoelectric equation is given by: 
?? max
=
h?? ?? - ?? 
where: 
?? max
 is the maximum kinetic energy of the electrons. 
h?? ?? is the energy of the incident photon. 
?? is the work function of the metal. 
For the initial light source of wavelength ?? , we know: 
2eV=
h?? ?? - 1eV 
Solving for 
h?? ?? gives: 
h?? ?? = 2eV+ 1eV= 3eV 
Now, if the wavelength is halved to 
?? 2
, the photon energy becomes: 
h?? ?? /2
=
2h?? ?? = 2× 3eV= 6eV 
The maximum kinetic energy for the new wavelength is then: 
?? max
'
= 6eV- 1eV= 5eV 
Therefore, the maximum kinetic energy of the ejected electrons when the surface is illuminated 
by a light source of wavelength 
?? 2
 is 5 eV . 
The correct answer is Option A. 
Q4: A sub-atomic particle of mass ????
-????
 ???? is moving with a velocity ?? .???? ×
????
?? ?? /?? . Under the matter wave consideration, the particle will behave closely like 
 (?? = ?? .???? × ????
-????
 ?? .?? ) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. X-rays 
B. Infra-red radiation 
C. Gamma rays 
D. Visible radiation 
Ans: A 
Solution: 
To find the matter wave behavior of the particle, we use the de Broglie wavelength formula: 
?? =
h
????
 
where: 
h= 6.63× 10
-34
 J · s is Planck's constant, 
?? = 10
-30
 kg is the mass of the particle, and 
?? = 2.21× 10
6
 m/s is its velocity. 
Follow these steps: 
Substitute the values into the formula: 
?? =
6.63× 10
-34
(10
-30
)(2.21× 10
6
)
 
Calculate the denominator: 
10
-30
× 2.21× 10
6
= 2.21× 10
-24
 
Now compute the wavelength: 
?? ˜
6.63× 10
-34
2.21× 10
-24
˜ 3 × 10
-10
 m 
This wavelength, roughly 3 × 10
-10
 m (or 0.3 nm ), lies within the X-ray portion of the 
electromagnetic spectrum, since X-rays typically have wavelengths in the range of about 0.01 
nm to 10 nm . 
Therefore, under the matter wave consideration, the particle will behave closely like: 
Option A: X-rays. 
Q5: In photoelectric effect an em-wave is incident on a metal surface and electrons 
are ejected from the surface. If the work function of the metal is 2.14 eV and stopping 
Page 4


JEE Main Previous Year Questions 
(2025): Dual Nature of Matter and 
Radiation 
Q1: An electron in the ground state of the hydrogen atom has the orbital radius of 
?? .?? × ????
-????
 ?? while that for the electron in third excited state is ?? .???? × ????
-????
 ?? . 
The ratio of the de Broglie wavelengths of electron in the ground state to that in the 
excited state is 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 4 
B. 3 
C. 9 
D. 16 
Ans: A 
Solution: 
We know, ?? =
h
????
 
and ?????? =
?? h
2?? 
? ???? =
?? h
2????
 
So, ?? =
h
?? h
2???? = 2?? ?? ?? 
? ?? ?
?? ?? 
We can write, 
?? ?? ?? ?? = (
?? ?? ?? ?? )(
?? ?? ?? ?? ) 
?
?? ?? ?? ?? = (
5.3× 10
-11
84.8× 10
-11
)(
4
1
) =
1
4
 
?
?? ?? ?? ?? = 4 
Q2: The work functions of cesium ( Cs ) and lithium ( Li ) metals are 1.9 eV and 2.5 eV , 
respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, 
then photo-electric effect is possible for the case of 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Both Cs and Li 
B. Neither Cs nor Li 
C. Li only 
D. Cs only 
Ans: D 
Solution: 
For photo-electric effect, energy of the photon must be greater than the work function of the 
metal. 
We know, ?? =
1240
?? =
1240
550
 (as ?? = 550 nm given) 
? ?? = 2.25eV 
here, ?? > ?? ????
 
So, ?? ?? only. 
Q3: A light source of wavelength ?? illuminates a metal surface and electrons are 
ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a 
light source of wavelength 
?? ?? , then the maximum kinetic energy of ejected electrons 
will be (The work function of metal is ?? ???? ) 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 5 eV 
B. 3 eV 
C. 2 eV 
D. 6 eV 
Ans: A 
Solution: 
Let's break down the problem step by step. 
The photoelectric equation is given by: 
?? max
=
h?? ?? - ?? 
where: 
?? max
 is the maximum kinetic energy of the electrons. 
h?? ?? is the energy of the incident photon. 
?? is the work function of the metal. 
For the initial light source of wavelength ?? , we know: 
2eV=
h?? ?? - 1eV 
Solving for 
h?? ?? gives: 
h?? ?? = 2eV+ 1eV= 3eV 
Now, if the wavelength is halved to 
?? 2
, the photon energy becomes: 
h?? ?? /2
=
2h?? ?? = 2× 3eV= 6eV 
The maximum kinetic energy for the new wavelength is then: 
?? max
'
= 6eV- 1eV= 5eV 
Therefore, the maximum kinetic energy of the ejected electrons when the surface is illuminated 
by a light source of wavelength 
?? 2
 is 5 eV . 
The correct answer is Option A. 
Q4: A sub-atomic particle of mass ????
-????
 ???? is moving with a velocity ?? .???? ×
????
?? ?? /?? . Under the matter wave consideration, the particle will behave closely like 
 (?? = ?? .???? × ????
-????
 ?? .?? ) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. X-rays 
B. Infra-red radiation 
C. Gamma rays 
D. Visible radiation 
Ans: A 
Solution: 
To find the matter wave behavior of the particle, we use the de Broglie wavelength formula: 
?? =
h
????
 
where: 
h= 6.63× 10
-34
 J · s is Planck's constant, 
?? = 10
-30
 kg is the mass of the particle, and 
?? = 2.21× 10
6
 m/s is its velocity. 
Follow these steps: 
Substitute the values into the formula: 
?? =
6.63× 10
-34
(10
-30
)(2.21× 10
6
)
 
Calculate the denominator: 
10
-30
× 2.21× 10
6
= 2.21× 10
-24
 
Now compute the wavelength: 
?? ˜
6.63× 10
-34
2.21× 10
-24
˜ 3 × 10
-10
 m 
This wavelength, roughly 3 × 10
-10
 m (or 0.3 nm ), lies within the X-ray portion of the 
electromagnetic spectrum, since X-rays typically have wavelengths in the range of about 0.01 
nm to 10 nm . 
Therefore, under the matter wave consideration, the particle will behave closely like: 
Option A: X-rays. 
Q5: In photoelectric effect an em-wave is incident on a metal surface and electrons 
are ejected from the surface. If the work function of the metal is 2.14 eV and stopping 
potential is 2 V , what is the wavelength of the em-wave?(Given ???? = ???????? ???????? 
where ?? is the Planck's constant and ?? is the speed of light in vaccum.) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 400 nm 
B. 600 nm 
C. 300 nm 
D. 200 nm 
Ans: C 
Solution: 
Let's break down the problem step by step: 
In the photoelectric effect, the energy of a photon is given by: 
h?? = ?? + ?? ?? 
where: 
?? is the work function of the metal, 
?? ?? is the kinetic energy of the ejected electron. 
The kinetic energy of the ejected electrons is provided by the stopping potential: 
?? ?? = ???? 
Given that the stopping potential is ?? = 2 V, we have: 
?? ?? = 2eV 
Now, we can calculate the photon energy: 
h?? = ?? + ?? ?? = 2.14eV+ 2eV= 4.14eV 
The relation between the photon's energy and its wavelength is: 
h?? =
h?? ?? 
Rearranging for ?? : 
?? =
h?? h?? 
Given that h?? = 1242eV· nm , substitute the values: 
?? =
1242eV· nm
4.14eV
 
Calculate the wavelength: 
?? ˜
1242
4.14
˜ 300 nm 
Thus, the wavelength of the incident electromagnetic wave is approximately 300 nm . 
The correct answer is Option C: 300 nm . 
Q6: An electron of mass ' ?? ' with an initial velocity ??? = ?? ?? ??ˆ(?? ?? > ?? ) enters an electric 
field ?? ? ? 
= -?? ?? ?? ˆ
. If the initial de Broglie wavelength is ?? ?? , the value after time ?? would 
be 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
Page 5


JEE Main Previous Year Questions 
(2025): Dual Nature of Matter and 
Radiation 
Q1: An electron in the ground state of the hydrogen atom has the orbital radius of 
?? .?? × ????
-????
 ?? while that for the electron in third excited state is ?? .???? × ????
-????
 ?? . 
The ratio of the de Broglie wavelengths of electron in the ground state to that in the 
excited state is 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 4 
B. 3 
C. 9 
D. 16 
Ans: A 
Solution: 
We know, ?? =
h
????
 
and ?????? =
?? h
2?? 
? ???? =
?? h
2????
 
So, ?? =
h
?? h
2???? = 2?? ?? ?? 
? ?? ?
?? ?? 
We can write, 
?? ?? ?? ?? = (
?? ?? ?? ?? )(
?? ?? ?? ?? ) 
?
?? ?? ?? ?? = (
5.3× 10
-11
84.8× 10
-11
)(
4
1
) =
1
4
 
?
?? ?? ?? ?? = 4 
Q2: The work functions of cesium ( Cs ) and lithium ( Li ) metals are 1.9 eV and 2.5 eV , 
respectively. If we incident a light of wavelength 550 nm on these two metal surfaces, 
then photo-electric effect is possible for the case of 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Both Cs and Li 
B. Neither Cs nor Li 
C. Li only 
D. Cs only 
Ans: D 
Solution: 
For photo-electric effect, energy of the photon must be greater than the work function of the 
metal. 
We know, ?? =
1240
?? =
1240
550
 (as ?? = 550 nm given) 
? ?? = 2.25eV 
here, ?? > ?? ????
 
So, ?? ?? only. 
Q3: A light source of wavelength ?? illuminates a metal surface and electrons are 
ejected with maximum kinetic energy of 2 eV . If the same surface is illuminated by a 
light source of wavelength 
?? ?? , then the maximum kinetic energy of ejected electrons 
will be (The work function of metal is ?? ???? ) 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 5 eV 
B. 3 eV 
C. 2 eV 
D. 6 eV 
Ans: A 
Solution: 
Let's break down the problem step by step. 
The photoelectric equation is given by: 
?? max
=
h?? ?? - ?? 
where: 
?? max
 is the maximum kinetic energy of the electrons. 
h?? ?? is the energy of the incident photon. 
?? is the work function of the metal. 
For the initial light source of wavelength ?? , we know: 
2eV=
h?? ?? - 1eV 
Solving for 
h?? ?? gives: 
h?? ?? = 2eV+ 1eV= 3eV 
Now, if the wavelength is halved to 
?? 2
, the photon energy becomes: 
h?? ?? /2
=
2h?? ?? = 2× 3eV= 6eV 
The maximum kinetic energy for the new wavelength is then: 
?? max
'
= 6eV- 1eV= 5eV 
Therefore, the maximum kinetic energy of the ejected electrons when the surface is illuminated 
by a light source of wavelength 
?? 2
 is 5 eV . 
The correct answer is Option A. 
Q4: A sub-atomic particle of mass ????
-????
 ???? is moving with a velocity ?? .???? ×
????
?? ?? /?? . Under the matter wave consideration, the particle will behave closely like 
 (?? = ?? .???? × ????
-????
 ?? .?? ) 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. X-rays 
B. Infra-red radiation 
C. Gamma rays 
D. Visible radiation 
Ans: A 
Solution: 
To find the matter wave behavior of the particle, we use the de Broglie wavelength formula: 
?? =
h
????
 
where: 
h= 6.63× 10
-34
 J · s is Planck's constant, 
?? = 10
-30
 kg is the mass of the particle, and 
?? = 2.21× 10
6
 m/s is its velocity. 
Follow these steps: 
Substitute the values into the formula: 
?? =
6.63× 10
-34
(10
-30
)(2.21× 10
6
)
 
Calculate the denominator: 
10
-30
× 2.21× 10
6
= 2.21× 10
-24
 
Now compute the wavelength: 
?? ˜
6.63× 10
-34
2.21× 10
-24
˜ 3 × 10
-10
 m 
This wavelength, roughly 3 × 10
-10
 m (or 0.3 nm ), lies within the X-ray portion of the 
electromagnetic spectrum, since X-rays typically have wavelengths in the range of about 0.01 
nm to 10 nm . 
Therefore, under the matter wave consideration, the particle will behave closely like: 
Option A: X-rays. 
Q5: In photoelectric effect an em-wave is incident on a metal surface and electrons 
are ejected from the surface. If the work function of the metal is 2.14 eV and stopping 
potential is 2 V , what is the wavelength of the em-wave?(Given ???? = ???????? ???????? 
where ?? is the Planck's constant and ?? is the speed of light in vaccum.) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. 400 nm 
B. 600 nm 
C. 300 nm 
D. 200 nm 
Ans: C 
Solution: 
Let's break down the problem step by step: 
In the photoelectric effect, the energy of a photon is given by: 
h?? = ?? + ?? ?? 
where: 
?? is the work function of the metal, 
?? ?? is the kinetic energy of the ejected electron. 
The kinetic energy of the ejected electrons is provided by the stopping potential: 
?? ?? = ???? 
Given that the stopping potential is ?? = 2 V, we have: 
?? ?? = 2eV 
Now, we can calculate the photon energy: 
h?? = ?? + ?? ?? = 2.14eV+ 2eV= 4.14eV 
The relation between the photon's energy and its wavelength is: 
h?? =
h?? ?? 
Rearranging for ?? : 
?? =
h?? h?? 
Given that h?? = 1242eV· nm , substitute the values: 
?? =
1242eV· nm
4.14eV
 
Calculate the wavelength: 
?? ˜
1242
4.14
˜ 300 nm 
Thus, the wavelength of the incident electromagnetic wave is approximately 300 nm . 
The correct answer is Option C: 300 nm . 
Q6: An electron of mass ' ?? ' with an initial velocity ??? = ?? ?? ??ˆ(?? ?? > ?? ) enters an electric 
field ?? ? ? 
= -?? ?? ?? ˆ
. If the initial de Broglie wavelength is ?? ?? , the value after time ?? would 
be 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 
?? ?? v1-
e
2
E
o
2
?? 2
 m
2
v
o
2
 
B. ?? 0
 
C. 
?? ?? v1+
e
2
E
o
2
t
2
 m
2
?? ?? 2
 
D. ?? o
v1+
e
2
E
o
2
t
2
 m
2
v
o
2
 
Ans: C 
Solution: 
Derivation of the de Broglie wavelength after time ?? 
The de Broglie wavelength is given by 
?? =
h
?? , 
where h is Planck's constant and ?? is the momentum of the particle. 
Initial Conditions: 
Initially, the electron has a velocity 
?? = ?? 0
??ˆ, 
so its momentum is 
?? 0
= ?? ?? 0
. 
The initial de Broglie wavelength is therefore 
?? 0
=
h
?? ?? 0
. 
Effect of the Electric Field: 
The electron enters an electric field 
?? ? 
= -?? 0
?? ˆ
. 
For an electron with charge -?? , the force is given by 
?? 
= -?? ?? ? 
= -?? (-?? 0
?? ˆ
)= ?? ?? 0
?? ˆ
 
This force causes an acceleration in the ?? ˆ
 direction: 
?? ?? =
?? ?? ?? =
?? ?? 0
?? . 
Momentum Components After Time ?? :
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FAQs on JEE Main Previous Year Questions (2025): Dual Nature of Matter and Radiation

1. What is the dual nature of matter and how does it relate to the concept of wave-particle duality?
Ans. The dual nature of matter refers to the concept that particles, such as electrons and photons, exhibit both wave-like and particle-like properties. This idea originated from the work of scientists like Louis de Broglie, who proposed that matter has wave characteristics, and Albert Einstein, who demonstrated the particle nature of light through the photoelectric effect. Wave-particle duality is fundamental in quantum mechanics, where entities like electrons can behave like waves, showing interference patterns, and also like particles, with defined positions and momenta.
2. What experiments demonstrate the dual nature of light?
Ans. Several key experiments illustrate the dual nature of light. The double-slit experiment is one of the most famous, where light passing through two closely spaced slits creates an interference pattern, characteristic of waves. However, when observed as particles (photons), light can also display discrete impacts on a detector, confirming its particle-like nature. Additionally, the photoelectric effect, where light ejects electrons from a metal surface, demonstrates that light behaves as a stream of particles (quanta or photons) with specific energy levels.
3. How did the discovery of the electron contribute to the understanding of the dual nature of matter?
Ans. The discovery of the electron by J.J. Thomson in 1897 was pivotal in understanding the dual nature of matter. Electrons were found to exhibit properties of both particles and waves. Thomson's experiments showed that electrons could be deflected by electric and magnetic fields, indicating their particle nature. Later, de Broglie's hypothesis that electrons could also behave as waves was confirmed through experiments such as electron diffraction, reinforcing the concept of wave-particle duality in fundamental particles.
4. What is the significance of the Heisenberg Uncertainty Principle in the context of wave-particle duality?
Ans. The Heisenberg Uncertainty Principle states that it is impossible to simultaneously know both the exact position and momentum of a particle. This principle is significant in the context of wave-particle duality because it emphasizes the limitations of classical mechanics when applied to quantum objects. It reflects the inherent uncertainty and probabilistic nature of particles like electrons, which can only be described in terms of wave functions, further illustrating their dual characteristics.
5. How does the dual nature of matter affect the field of quantum mechanics?
Ans. The dual nature of matter is a cornerstone of quantum mechanics, fundamentally altering our understanding of physical phenomena. It leads to the development of quantum theories that describe the behavior of particles at microscopic scales, including the principles of superposition and entanglement. This duality necessitates the use of wave functions to predict probabilities of finding particles in certain states, rather than definitive outcomes, thereby reshaping the theoretical framework of physics and influencing technologies such as semiconductors and quantum computing.
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