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Modern Physics is a very important constituent of the Physics portion of JEE Main. Modern Physics for JEE Main is based on two branches: Relativity and Quantum Mechanics. It is the most scoring part of the JEE Main. Some of the topics are Nuclear Fission and Fusion, Photoelectric effect, Plank’s Theory, etc. These topics are quite fascinating but involve concepts that must be understood properly.

- A minimum of 4-6 questions will be asked in the JEE Main Paper from Modern Physics.
**Check JEE Main Physics Syllabus** - The weightage of the marks of Modern Physics in JEE Mains is 20 marks.
- Some important topics from Modern Physics are Photoelectricity, Electromagnetic Induction, Plan Theory, Atomic Models, etc.

Below are given some of the detailed topics along with the questions based on Modern Physics for **JEE Main** point of view.

**Must Read:**

Nuclear Fission and Fusion are the two different kinds of energy-releasing reactions. In these reactions, the energy is released from high- powered atomic bonds between the particles that are present in the nucleus. Both of these processes are opposite in nature. Fission involves the splitting of an atom into two or more atoms whereas two or more atoms combine to form a larger atom in the Fusion process.

The process of splitting a big atom into two or smaller atoms is termed as Nuclear Fission. It is a process in which the nucleus of an atom breaks down to form smaller nuclei along with the by-products that comprise of free neutrons and photons in the form of gamma rays, alpha and beta particles.

It is an exothermic reaction which means there is a release of a huge amount of energy when it takes place. This energy is used for nuclear power or maybe for the explosion of nuclear weapons.

**Question: In which of the following process are Neutrons emitted?**

- Inverse beta Decay
- Nuclear fission
- Spontaneous Fission
- Nuclear fusion

**Answer:** (2)

**Explanation:** In the Nuclear fission process a heavy nucleus is split into two or lighter nuclei. This results in a decrease in mass and consequent exothermic energy and emission of neutrons take place. Two to three neutrons are emitted per nuclei which are known as fission elements.

The process in which two or more atoms combine together to form a large atom is called Nuclear Fusion. It may also be defined as the process in which multiple nuclei join together to form a heavy nucleus. Nuclear Fusion is accompanied by the release or absorption of energy that depends on the masses of the nuclei involved.

**Question: Fusion reactions are called __________?**

- Thermonuclear
- Thermoduric
- Thermo Uric
- Compound reactions

**Answer:** (1)

**Explanation:** Fusion reactions are called thermonuclear because of the higher temperature required to trigger and sustain the reaction.

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- The positive charges and most of the mass of an atom is uniformly distributed over the full size of an atom.
- Electrons are studded in this Uniform distribution.
- The model is also known as the Plum Pudding model.

- Most of the mass of an atom and all the positive charges are concentrated within a size of 10-14 m inside the atom. This concentration is called the atomic nucleus.
- The electron revolves around the nucleus in circular orbits.

- Bohr improvised Rutherford atomic model and added some postulates in it.
- The electron does not radiate energy in a stable orbit.
- A stable orbit is that in which the angular momentum of the electron about the nucleus is an integral (n) multiple of h/2π, i.e., mvr = nh/2π; n = 1, 2, 3, ……. (n ≠ 0).
- If the electron jumps from a lower to a higher orbit or falls from a higher to a lower orbit. Then the electron can absorb or radiate energy only
- The energy emitted or absorbed is a light photon of frequency v and of energy, i.e., E = hv.

**Question: The correct set of four quantum numbers of the valence electrons of the rubidium atom is____?**

- 5, 0, 0, +½
- 5, 1, 0, + ½
- 5, 1, 1, +½
- 5, 0 , 1, +½

**Solution:** (1)

**Explanation:** Atomic Number of rubidium is Z = 37. Therefore its electronic configuration is [Kr] 5 s1. Since the valence electron enters in 5s subshell so the quantum numbers are n= 5, l = 0. As the value of m is from +l to -l so m = o and s = ½ or -½

**The wave properties are exhibited by beams of electrons and other forms of matter that includes interference and diffraction and has a De- Broglie wavelength that is given by:**

**⋋=hp=hmv**

Where ⋋denotes the wavelength of the particle,

h stands for Plank’s constant,

m denotes the mass of the particle,

v is the velocity of the particle.

**Question: An photon and an electron are having the same wavelength. If p is the momentum of the electron and E is the energy of the photon. The magnitude of p/E in S.I unit is**

- 3.010
^{8} - 3.3310
^{-9} - 9.110
^{-31} - 6.6410
^{34}

**Solution:** (2)

**Explanation:** As we know that, for electron, λ = h/p

or, p = h/λ

and for photon E = hc/λ

Thus, p/E = 1/c = 1/(310** ^{8}** m/s)

= 3.3310** ^{-9}** s/m

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The phenomenon of the emission of electrons, when metals are exposed to light at a certain minimum frequency is called the Photoelectric effect.

- The concept of the Photoelectric effect can only be explained on the basis of the quantum theory.
- It is an instantaneous process.
- The electrons are emitted only if the incident light has the frequency v ≥ v0 (Threshold frequency) and the emission of electrons is independent of intensity.
- The wavelength corresponding to v0 is called threshold wavelength λ0 .
- Also, the number of electrons emitted per second depends on the intensity of the incident light.
- The maximum velocity and the stopping potential of the electrons depends upon the frequency of incident light and is independent of its frequency.

**Einstein gave the Photoelectric equation which is stated as:**

- Photon energy = K. E. of electron + work function.
- h v = 1/2 mv
+ ϕ^{2}

where ϕ = Work function = energy needed by the electron in freeing itself from the atoms of the metal, i.e., ϕ = h v0

**Question: Concerning the photoelectric effect, which of the following is not true?**

- Ultraviolet light is needed for the photoelectric effect to occur for most of the metals
- It takes a few minutes before electrons are emitted from the metal it is shining upon because a faint light contains very little energy
- Bright light causes more electrons to be emitted than a faint light.
- Higher frequency light emits electrons with higher kinetic energies.

**Solution:** (2)

The emission spectrum of atomic hydrogen has been divided into a number of spectral series. The wavelengths of these are given by the Rydberg formula. The observed spectral lines are due to the transitions between two energy levels in an atom by electrons.

**Lyman Series:**(Landing Orbit n= 1)

Ultraviolet Region: V- =R112-1n22; n2 > 1

**Balmer Series:**(Landing Orbit n= 2)

Visible Region: V- =R122-1n22; n2 > 2

**Paschen Series:**(Landing Orbit n= 3)

In the near Infrared Region: V- =R132-1n22; n2 > 3

**Bracket Series:**(Landing Orbit n= 4)

In the mid Infrared Region: V- =R142-1n22; n2 > 4

**Pfund Series:**(Landing Orbit n= 5)

In the far Infrared Region: V- =R152-1n22; n2 > 5

In all the series n2 = n1 + 1 is the α line

= n1 + 2 is the β line

= n1 + 3 is the γ line ……. etc .

where n1 = Landing orbit

**Question: Balmer series lies in_________?**

- visible region
- infrared region
- ultraviolet region
- none of the above

**Solution:** (1)

Radioactive decay is the process that refers to the decay to attain stability. It simply refers to the loss of particles from an unstable atom in order to attain more stability. The unstable elements emit some particles from their nucleus to gain stability and this process is termed as radioactivity. Uniformity is produced when we have an equal number of neutrons and protons in elements that determine and directs the nuclear forces to keep the nuclear particles inside the nucleus. There are chances that when a particle becomes more frequent than another and creates an unstable nucleus. This unstable nucleus can release radiation so that it can gain stability.

**Radioactive decay can occur in five different forms:**

- Alpha emission
- Beta emission
- Positron emission
- Electron capture
- Gamma emission

**Question: N1 atoms of a radioactive element emit N2 beta particles per second. The decay constant of the element is (in s-1)**

- N1/N2
- N2/N1
- N1ln(2)
- N2 ln(2)

**Solution: **(2)

**Explanation:** Activity of a radioactive substance R = λN

So, λ = R/N

Here, R = N2 and N = N1

Thus, λ = N2/N1

**Check Gujarat Board Released Question Banks for JEE Main**

**1. Question: If an alpha particle of energy 5 MeV is scattered and the scattering is through 180° by a fixed uranium nucleus. Then calculate the distance of the closest approach is of what order?**

- 1A
^{0} - 10
^{-10} - 10
^{-12} - 10
^{-15}

**Tip to Solve the Question:** Keep in mind that energy is conserved and Loss in Kinetic Energy = Loss in Potential Energy

**Solution:** (3)

**2. Question: The half-life of radioactive radon is 3.8 days. Calculate what will be the time at the end of which (l/20)th of the radon sample will remain undecayed? Also given log10e= 0.4343**

- 8 Days
- 16.5 Dyas
- 33 Days
- 76 Days

**Tip to Solve the Question:** Here, first of all, you have to calculate the value of “⋋”. Then after calculating this value you can calculate the time that needs to convert in the form of days.

**Solution:** (2)

**3. Question: What will be the number of possible elements if elements with principal quantum number 4 were not allowed in nature?**

- 60
- 32
- 4
- 64

**Tip to Solve the Question:** Calculate by considering that the minimum of electrons in an orbit are 2n^{2}

**Solution:** (1)

- Calculate the wavelength first wherever necessary.
- Most of the questions from this section will include calculating the electrons emission with respect to the time. So in these questions, you have to consider the time factor from the start of the question.
- Do not get confused with the long statements of the questions. Most of the questions will be direct but include the unnecessary data that you have to identify to calculate the answer quickly.
- If in some questions you have to calculate p/E then start from the formula λ = h/p and E = hc/λ. From these two formulas, you can calculate the p/E value

- Students must go through the detailed syllabus and divide each topic according to the time left to prepare for the examination.
- Class Notes are the perfect example of an initiating step that helps you begin from scratch and build your concepts strong slowly. Always remember to pen down all the important information you receive in the class. Class notes will provide the basic foundation steps
**JEE Main 2020**. Once the basic concepts are clear and understood, students can move on to build on this knowledge. - Candidates can purchase the online test series by some of the best and top coaching institutes in online mode only. The tests will be a dummy of the actual question paper and will definitely help the candidates in getting an overview of the actual question paper and prepare for it in a better way.
- Online tutorials are a good source of preparation as one can find unlimited video lectures on each subject. The content offered through these channels is reliable, given by field experts. Also, students can rely on these video lectures in case they’ve missed a topic or had doubts even after attending the lecture.
- Candidates are advised to attempt at least 10- 15 previous year sample papers before appearing for the actual examination to understand the paper pattern, marking scheme, and types of questions asked in the examination.

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