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NATIONAL LEVEL ONLINE TEST
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.
Below are given some of the detailed topics along with the questions based on Modern Physics for JEE Main point of view.
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?
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 __________?
Explanation: Fusion reactions are called thermonuclear because of the higher temperature required to trigger and sustain the reaction.
Question: The correct set of four quantum numbers of the valence electrons of the rubidium atom is____?
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:
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
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/(3108 m/s)
= 3.3310-9 s/m
The phenomenon of the emission of electrons, when metals are exposed to light at a certain minimum frequency is called the Photoelectric effect.
Einstein gave the Photoelectric equation which is stated as:
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?
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.
Ultraviolet Region: V- =R112-1n22; n2 > 1
Visible Region: V- =R122-1n22; n2 > 2
In the near Infrared Region: V- =R132-1n22; n2 > 3
In the mid Infrared Region: V- =R142-1n22; n2 > 4
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_________?
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:
Question: N1 atoms of a radioactive element emit N2 beta particles per second. The decay constant of the element is (in s-1)
Explanation: Activity of a radioactive substance R = λN
So, λ = R/N
Here, R = N2 and N = N1
Thus, λ = N2/N1
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?
Tip to Solve the Question: Keep in mind that energy is conserved and Loss in Kinetic Energy = Loss in Potential Energy
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
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.
3. Question: What will be the number of possible elements if elements with principal quantum number 4 were not allowed in nature?
Tip to Solve the Question: Calculate by considering that the minimum of electrons in an orbit are 2n2