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Modern Physics - Physics JEE Advanced

Q: 4. What are the key topics covered in Modern Physics for JEE Advanced?

Key topics include:Quantum MechanicsPhotoelectric EffectBohr's Atomic ModelDual Nature of Matter and RadiationNuclear PhysicsSemiconductors

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Physics

Modern Physics

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Concepts of Modern Physics for JEE Advanced Physics

Let us first understand what is modern physics? Modern Physics is a branch of physics that deals with the fundamental nature of the universe with post-Newtonian concepts. Galileo Galilei, who is also known as the father of modern physics, contributed significantly in modern physics. In the early twentieth century, some experimental results could not be matched with the predictions of classical physics, which describes physical phenomena at an ordinary scale. Modern physics gradually took birth from these theories.

In JEE, this unit is the most important unit of physics. This unit contains three chapters from the class 12 syllabus, namely, Dual Nature of Matter, Atoms and Nuclei. In this article we will cover the important topics of all the three chapters and understand the elements of modern physics.

In this article, we will be focusing on the concepts of Modern Physics that play a vital role in the JEE exam.

Important Topics of Modern Physics Chapter

Photoelectric effect
Cathode rays
De Broglie equation
Rutherford’s Model of Atom
Bohr Model
Hydrogen Spectrum
Composition and Size of Nucleus
Radioactivity
Nuclear Fission and Fusion.

Modern Physics Important Concept for JEE Advanced

We know that in a chapter there will be many concepts that are being mentioned in the textbook, but when it comes to JEE preparation we have to focus on what is most important among all the concepts in the chapter. Below is the table provided for some most important concepts of Modern Physics.

Sl. No	Name of the Concept	Key Points of Concept
	Photoelectric effect	The phenomenon of the photoelectric effect revolves around electron emission. The photoelectric effect shows that when light radiation is incident on a metallic surface, the electrons near the surface absorb the radiation energy and get liberated from the surface. These photo or light generated electrons are known as photoelectrons.
	Relation between Potential and Photoelectric Current	With an increase in the accelerating potential, the photoelectric current increases equally. The saturation current occurs when the photoelectric current reaches its maximum value. All photoelectrons are emitted from the emitter and reach the collection plate when the photoelectrons reach saturation.
	Relation between Intensity of the Incident Light and Photoelectric Current	Photocurrent does not occur if the frequency of the incident radiation or light is lower than the metal surface's minimum frequency. It rapidly declines until it reaches zero, which is also known as the cut-off frequency or the metal plate's stopping potential. The photoelectric current's cut-off frequency, or stopping potential, is essentially a physical feature of the metal plate. The photocurrent becomes zero or reaches the cut-off frequency only when the stopping potential is sufficient to drive back even the highest-energy photoelectric particles approaching with maximum kinetic energy.
4	Cathode rays	Cathode rays are streams of fast-moving electrons emitted from the cathode (negative electrode) of a vacuum tube when a high voltage is applied between the cathode and the anode (positive electrode). Properties of cathode rays: Cathode rays travel in straight lines and can cast sharp shadows. Cathode rays are deflected by electric and magnetic fields. Cathode rays have mass and energy. Cathode rays can produce heat, light, and X-rays. Cathode rays can ionize gases.
5	De Broglie hypothesis- dual nature of light	De Broglie theory suggests that light behaves like both particles and waves. According to De Broglie's hypothesis, the wavelength associated with momentum is too small, which explains why microscopic elements do not exhibit wave characteristics.
6	Rutherfor’s atomic model and its limitation	Every atom consists of a tiny central core, called the atomic nucleus in which the entire positive charge and almost entire mass of the atom are concentrated. The atom as a whole is electrically neutral and electrons revolve around the nucleus in various circular orbits. As the electron revolves so it must lose energy continuously, it must spiral inwards and eventually fall into the nucleus. Therefore, atoms should emit a continuous spectrum but what we observe is a line spectrum. So this point isn’t explained by the Rutherford model which is the biggest limitation of this model.
7	Bohr’s model of hydrogen atom	Bohr gives three postulates; Every atom consists of a central core called nucleus which contains the positive charge and electrons revolve around it in their circular orbits. Electrons can revolve only in certain discrete non radiating orbits, called stationary orbits for which total angular momentum of the revolving electron is an integral multiple of $\dfrac{h}{2\pi}$, where $h$ is the Planck’s constant. The emission or absorption of energy occurs only when an electron jumps from one of its specified orbit to another. The difference in the total energy of electrons in the two permitted orbits is absorbed when the electron jumps from inner to the outer orbit.
8	Hydrogen Spectrum	The hydrogen spectrum is the line spectrum of the hydrogen atom. It is a series of discrete wavelengths of electromagnetic radiation that is emitted or absorbed by hydrogen atoms. The hydrogen spectrum is important because it provides evidence for the quantized energy levels of the hydrogen atom.
9	Series of hydrogen spectrum	Lyman series: This series corresponds to transitions from higher energy levels to the first energy level. The wavelengths of the Lyman series lines are in the ultraviolet region of the spectrum. Balmer series: This series corresponds to transitions from higher energy levels to the second energy level. The wavelengths of the Balmer series lines are in the visible region of the spectrum. Paschen series: This series corresponds to transitions from higher energy levels to the third energy level. The wavelengths of the Paschen series lines are in the infrared region of the spectrum. Brackett series: This series corresponds to transitions from higher energy levels to the fourth energy level. The wavelengths of the Brackett series lines are in the far infrared region of the spectrum. Pfund series: This series corresponds to transitions from higher energy levels to the fifth energy level. The wavelengths of the Pfund series lines are in the far infrared region of the spectrum.
10	Nucleus of an Atom	The nucleus of an atom is a small, dense region at the center of the atom that contains the majority of the atom's mass and positive charge. The nucleus is composed of two types of particles: protons and neutrons. Protons have a positive charge and neutrons have no charge. The number of protons in the nucleus is called the atomic number, and it is what determines the identity of an element. The number of neutrons in the nucleus is called the neutron number, and it can vary for different isotopes of the same element. The nucleus of an atom is very small compared to the size of the atom as a whole. If the atom were expanded to the size of a football stadium, the nucleus would be about the size of a marble.
11	Composition of the Nucleus	The nucleus of an atom is made up of protons and neutrons. Protons have a positive charge, and neutrons have no charge. The number of protons in the nucleus is called the atomic number, and it is what determines the identity of an element. The number of neutrons in the nucleus is called the neutron number, and it can vary for different isotopes of the same element.
12	Size of the Nucleus	The nucleus of an atom is very small compared to the size of the atom as a whole. The radius of a nucleus is about 100,000 times smaller than the radius of the atom. This means that if the atom were expanded to the size of a football stadium, the nucleus would be about the size of a marble. The small size of the nucleus is due to the strong nuclear force, which holds the protons and neutrons together. The strong nuclear force is one of the four fundamental forces in nature, and it is much stronger than the electromagnetic force that holds electrons in orbit around the nucleus.
13	Radioactivity	Radioactivity is the process by virtue of which a heavy element disintegrates itself without being forced by any external agent to do so. According to laws of radioactive disintegration, the number of atoms disintegrating per second at any instant is directly proportional to the number of radioactive atoms actually present in the sample at that instant. $-\dfrac{dN}{dt}\varpropto N$ The half life of a radioactive element is the amount of time it takes for half of the atoms present in the sample to decay. Average life time of the element is obtained by calculating the total life time of all the atoms of the element and dividing it by the total number of atoms present initially in the sample of the element
14	Alpha, beta and Gamma decay	Alpha decay is the phenomenon of emission of an alpha particle from a radioactive nucleus. Ex: $U^{238}_{92}\longrightarrow Th^{234}_{90}+He^4_2$ Beta decay is the phenomenon of emission of an electron from a radioactive nucleus. Ex: $Th^{234}_{90}\longrightarrow Pa^{234}_{91}+e^0_{-1}$ Gamma decay is the phenomenon of emission of gamma ray photons from a radioactive nucleus. Ex: $Ni^{60*}_{28}\longrightarrow Ni^{60}_{27}+E_{\gamma}$
15	Nuclear fission and fusion	Nuclear fission is the process of splitting a heavy nucleus (usually A>230) into two or more lighter nuclei. Nuclear fusion is the process by which two or more lighter nuclei combine to produce a single heavy nucleus.

Check out more details about Modern Physics, its concepts and relativity.

List of Important Formulas for Modern Physics Chapter

Sl. No	Name of Concept	Formula
1.	De Broglie wavelength	$\lambda=\dfrac {h}{p}$ Or $\lambda=\dfrac {h}{\sqrt {2mE}}$
2.	Photon energy	$E=h\nu=\dfrac {hc}{\lambda}$
3.	Einstein’s photoelectric equation and Work function	${E_k}=h\nu-\phi$ Where, $\phi$- The work function
4.	Threshold frequency	${\nu_o}=\dfrac {\phi}{h}$
5.	Momentum of photon	The energy of photon moving with momentum p is $E=p c$ Therefore, momentum p is given by: $p=\dfrac {h \nu}{c}$
6.	Bohr’s model of hydrogen atom	From 1st postulate of Bohr’s model of atom we can write; $\dfrac{mv^2}{r}=\dfrac{KZe^2}{r^2}$ Here, $v$= velocity with which electron is moving and $r$= radius of the orbit. From 2nd postulate, $mvr=\dfrac{nh}{2\pi}$ From 3rd postulate, $E_2-E_1=nh\nu$ The electron's frequency in Bohr's stationary orbit is, $\nu=\dfrac{KZe^2}{nhr}$ Total energy of electron in Bohr’s stationary orbit is, $E=-\dfrac{13.6}{n^2}eV$
7.	Radioactivity	According to the law of radioactive decay, $-\dfrac{dN}{dt}\varpropto N$ $R=-\dfrac{dN}{dt}= \lambda N$ Here, $\lambda$ =disintegration constant. Half life ($T_{1/2}$)of radioactive element is, $T_{1/2}=\dfrac{0.693}{\lambda}$ Average life ($\tau$) is, $\tau= \dfrac{1}{\lambda}$ The activity of an element is given as, $A=A_o e^{-\lambda t}$

JEE Advanced Modern Physics Solved Examples

1. Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work-function for sodium is 1.82 eV.Find

(a) the energy of the photons causing the photoelectrons emission.

(b) the quantum numbers of the two levels involved in the emission of these photons.

(c) the change in the angular momentum of the electron in the hydrogen atom, in the above transition (Ionization potential of hydrogen is 13.6 eV.)

Sol:

Given,

Maximum Kinetic Energy for Sodium (KEmax) = 0.73 eV,

Work-function for sodium (W) = 1.82 eV,

(a) The energy of photons:

From Einstein’s equation of photoelectric effect,

Energy of photons (E) causing the photoelectric emission = Maximum kinetic energy of emitted photons (KEmax) + work-function (W)

E = KEmax + W = 0.73 eV + 1.82 eV = 2.55 eV,

(b) For Hydrogen atom,

E1 = - 13.6 eV, E2 = 3.4 eV, E3 = -1.5 eV, E4 = - 0.85 eV

Now, from the levels given above, E2 - E4 = 2.55 eV

Therefore, quantum numbers of the two levels involved in the emission of these photons are 4 and 2.

Here, angular momentum (L) is given by $\dfrac{nh}{2\pi}$, where, n = quantum number and h = planck’s constant,

Therefore, $\Delta L=L_2-L_4 = \dfrac{2h}{2\pi}-\dfrac{4h}{2\pi}$

Hence, Change in angular momentum will be $-\dfrac{h}{\pi}$

Key Point: Kinetic energy of the fastest moving electron is maximum kinetic energy only because kinetic energy depends on mass and velocity only.

2. The mean lives of an unstable nucleus in two different decay processes are 1620 yr and 405 yr, respectively. Find out the time during which three-fourth of a sample will decay.

Sol:

Given,

Mean life of Process- 1 (t1) = 1620 yr,

Mean life of Process- 2 (t2) = 405 yr,

Let decay constants of process-1 and process- 2 be λ1 and λ2, respectively. Then,

$\lambda_1 = \dfrac{1}{t_1}$

$\lambda_2 = \dfrac{1}{t_2}$

If the effective decay constant is λ, then

λN = λ1N + λ2N

λ = λ1 + λ2

$\lambda = \dfrac{1}{t_1}+\dfrac{1}{t_2}$

$\lambda = \dfrac{1}{1620}+\dfrac{1}{405}$ year-1

$\lambda = \dfrac{1}{324}$ year-1

Now, when three- fourth of the sample will decay. The sample remaining will be one- fourth of the total sample. If the initial sample was N0, then final sample will have N0/4,

So, $\dfrac{N_0}{4} = N_0e^{-\lambda t}$

Applying logarithms on both the sides, we get,

$-\lambda t = \ln\lgroup\dfrac{1}{4}\rgroup$ = -1.386

t = 449 yr

Therefore, the time during which three-fourth of a sample will decay will be 449 yr.

Key point: We need to modify the formula according to the given data.

Previous Year Questions From Modern Physics

1. The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3 \times {10^{-3}}$ watt will be: (JEE 2021)

${10^{16}}$
${10^{15}}$
${10^{18}}$
${10^{17}}$

Sol:

Given,

Wavelength of source of monochromatic light=$\lambda$= 600 nm

Power radiated by the source=$3.3 \times {10^{-3}}$ watt

The energy of the photon=E=$\dfrac {hc}{\lambda}= 3.3 \times {10^{-19}} Joules$

Let n is the number of photons emitted per second such that the energy emitted in one second is $3.3 \times {10^{-3}}$ J.

Thus, we get:

nE=$3.3 \times {10^{-3}}$

$n=\dfrac {3.3 \times {10^{-3}}}{3.3 \times {10^{-19}}}={10^{16}}$

Therefore, option A is the right answer.

Trick: Use the formula $P=\dfrac {nhc}{\lambda}$ to arrive at the solution easily.

2. There are $10^{10}$ radioactive nuclei in a given radioactive element, its half-life time is 1 minute. How many nuclei will remain after 30 seconds? (Take, $\sqrt{2}=1.414$) (JEE 2021)

a. $7 \times 10^{9}$

b. $2 \times 10^{9}$

c. $4 \times 10^{10}$

d. $10^{5}$

Sol:

Given that,

Original number of nuclei, ${N_o} = 10^{10}$

Half- life time, $t_{½} = 60\,s$

To find: Number of nuclei remain after time ($t$ = 30 seconds) i.e, $N$.

To solve this problem we have to use the concept of half life and also the relation of the number of nuclei decayed to the original number of nuclei present in the sample of a radioactive element.

Now using the concept of half live, we can write the relation between the number of nuclei decayed ($N$) to the original number of nuclei($N_o$) as,

$\dfrac{N}{N_o}=(\dfrac{1}{2})^{t/t_{1/2}}$

Now after putting the values of the quantities in the above relation, we get;

$\dfrac{N}{10^{10}}=(\dfrac{1}{2})^{30/60}$

$\dfrac{N}{10^{10}}=(\dfrac{1}{2})^{1/2}$

$N= \dfrac{10^{10}}{\sqrt{2}} \approx 7 \times 10^{9}$

Hence, the number of nuclei that remain after 30 seconds is $7 \times 10^{9}$.

Therefore, option a is correct.

Trick: The application of the half life concept and the relation between the number of nuclei decayed to the original number of nuclei in a sample is important to solve this problem.

Practice Questions

1. Wavelengths belonging to the Balmer series for hydrogen atoms lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work-function is 2.0 eV. Find (in eV) the maximum kinetic energy of the emitted photoelectrons.

(Ans: 0.55 eV)

2. A stable nuclei C is formed from two radioactive nuclei A and B with decay constant of 𝜆1 and 𝜆2 respectively. Initially, the number of nuclei of A is N0 and that of B is zero. Nuclei B are produced at a constant rate of P. Find the number of the nuclei of C after time t

(Ans:$N_c = N_0 (1-e^{-\lambda_1 t})+P \lgroup t+\dfrac{e^{-\lambda_2 t}-1}{\lambda_2}\rgroup$)

JEE Advanced Physics Modern Physics Study Materials

Here, you'll find a comprehensive collection of study resources for Modern Physics designed to help you excel in your JEE Advanced preparation. These materials cover various topics, providing you with a range of valuable content to support your studies. Simply click on the links below to access the study materials of Modern Physics and enhance your preparation for this challenging exam.

JEE Advanced Modern Physics Study Materials

JEE Advanced Modern Physics Notes

JEE Advanced Modern Physics Important Questions

JEE Advanced Modern Physics Practice Paper

JEE Advanced Physics Study and Practice Materials

Explore an array of resources in the JEE Advanced Physics Study and Practice Materials section. Our practice materials offer a wide variety of questions, comprehensive solutions, and a realistic test experience to elevate your preparation for the JEE Advanced exam. These tools are indispensable for self-assessment, boosting confidence, and refining problem-solving abilities, guaranteeing your readiness for the test. Explore the links below to enrich your Physics preparation.

JEE Advanced Physics Study and Practice Materials

JEE Advanced Physics Previous Year Question Papers

JEE Advanced Physics Formula

JEE Advanced Sample Paper

JEE Advanced Physics Difference Between

JEE Advanced Physics Chapters 2025

The JEE Advanced Physics syllabus for 2025 emphasizes in-depth understanding and application skills, crucial for the competitive engineering entrance exam.

JEE Advanced Physics Chapters
1	General Physics	5	Electromagnetic Waves
2	Mechanics	6	Optics
3	Thermal Physics	7	Modern Physics
4	Electricity and Magnestism

Conclusion

The Modern Physics chapter for JEE Advanced Physics is crucial for your preparation. It covers key topics like atoms, nuclei, and semiconductors, which are fundamental in understanding advanced physics concepts. It's important to focus on understanding the basic principles and theories thoroughly. Additionally, practicing numerical problems related to these topics is essential to build problem-solving skills. Pay attention to concepts like photoelectric effect, Bohr's model, and nuclear reactions as they often appear in exams. Lastly, make sure to revise regularly and clarify any doubts you may have. With a strong grasp of Modern Physics, you'll be well-prepared for the JEE Advanced exam.

FAQs on Modern Physics - Physics JEE Advanced

1. How many questions come from Modern Physics in JEE?

Modern Physics is having high importance in the JEE exam, it carries nearly 3-4 questions with a weightage 11%.

2. Which chapters are a part of Modern Physics?

This unit contains three chapters from the class 12 syllabus, namely, Dual Nature of Matter, Atoms and Nuclei.

3. Which chapter has more weightage in JEE?

Modern Physics, from the class 12th syllabus, is the most important topic for JEE. Current Electricity & Magnetism, Electrostatics and Optics are also very important chapters for JEE.

4. What are the key topics covered in Modern Physics for JEE Advanced?

Key topics include:

Quantum Mechanics
Photoelectric Effect
Bohr's Atomic Model
Dual Nature of Matter and Radiation
Nuclear Physics
Semiconductors

5. Why is Modern Physics important for JEE Advanced

Modern Physics constitutes a significant portion of the JEE Advanced syllabus and often carries a considerable weightage in the exam. Understanding these concepts is crucial for solving complex problems in various areas of physics.

6. What are some common misconceptions students have about Modern Physics?

Some students may find concepts like wave-particle duality or quantum mechanics difficult to grasp initially. They may also struggle with visualizing atomic and subatomic processes due to their abstract nature.

7. How can I improve my understanding of concepts like quantum mechanics and relativity in Modern Physics?

Practice is key. Solve numerous problems and experiment with thought experiments related to these concepts. Visual aids, simulations, and analogies can also help in understanding abstract concepts.

8. How should I approach solving numerical problems in Modern Physics?

Understand the given problem thoroughly, identify the relevant concepts and equations, and then proceed with solving step by step. Pay attention to units and always cross-check your answers.

9. Are there any specific resources or study materials recommended for mastering Modern Physics for JEE Advanced?

Vedantu's study materials, NCERT textbooks, previous years' question papers, and reference books like "Concepts of Modern Physics" by Arthur Beiser are highly recommended. Online resources such as video lectures and interactive simulations can also be beneficial.