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800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.
PhysicsClass 122 marksmedium
Atoms
State the main conclusions of Rutherford's alpha-particle scattering experiment.
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From the scattering of alpha particles by a thin gold foil, Rutherford concluded that: (1) most of the atom is empty space, since most alpha particles passed straight through; (2) the entire positive charge and almost all the mass of the atom are concentrated in a very small central region called the nucleus, since a few alpha particles were deflected through large angles; and (3) the electrons revolve around the nucleus, and the size of the nucleus is very small compared with the size of the atom.
Marking-scheme points
- ✓Most of the atom is empty space
- ✓Positive charge and mass concentrated in a tiny nucleus
- ✓Electrons revolve around the nucleus
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Atoms
State the postulates of Bohr's model of the hydrogen atom.
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Bohr's postulates are: (1) The electron revolves around the nucleus only in certain fixed circular orbits called stationary states, in which it does not radiate energy. (2) Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of h/(2 pi), that is, m v r = n h/(2 pi) (quantisation of angular momentum). (3) Energy is emitted or absorbed only when the electron jumps from one orbit to another, the energy of the emitted or absorbed photon being h f = E2 - E1.
m v r = n h/(2 pi)
Marking-scheme points
- ✓Electrons revolve in fixed stationary orbits without radiating
- ✓Angular momentum quantised: m v r = n h/(2 pi)
- ✓Energy change on jump: h f = E2 - E1
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Atoms
The energy of an electron in the ground state of hydrogen is -13.6 eV. Calculate the energy of the electron in the second orbit (n = 2).
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The energy of the electron in the nth orbit of hydrogen is En = -13.6/n^2 eV. For n = 2, E2 = -13.6/2^2 = -13.6/4 = -3.4 eV. The negative sign shows that the electron is bound to the nucleus, and the energy increases (becomes less negative) as n increases.
En = -13.6/n^2 eV
Marking-scheme points
- ✓En = -13.6/n^2 eV
- ✓E2 = -13.6/4
- ✓E2 = -3.4 eV
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Atoms
Calculate the energy of the photon emitted when an electron in a hydrogen atom jumps from n = 3 to n = 2. (Use En = -13.6/n^2 eV)
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Energy of the electron in n = 3: E3 = -13.6/9 = -1.51 eV. Energy in n = 2: E2 = -13.6/4 = -3.4 eV. The energy of the emitted photon = E3 - E2 = -1.51 - (-3.4) = 1.89 eV. This corresponds to the H-alpha line of the Balmer series (visible red light).
E(photon) = E(higher) - E(lower)
Marking-scheme points
- ✓E3 = -13.6/9 = -1.51 eV; E2 = -13.6/4 = -3.4 eV
- ✓Photon energy = E3 - E2 = 1.89 eV
- ✓This is the H-alpha (Balmer series) line
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Atoms
State two limitations of Bohr's model of the atom.
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(1) Bohr's model applies successfully only to hydrogen and hydrogen-like single-electron atoms; it fails to explain the spectra of atoms having more than one electron. (2) It could not explain the fine structure of spectral lines or the relative intensities of the lines, and it does not account for the splitting of spectral lines in electric and magnetic fields (the Stark and Zeeman effects). Also, it arbitrarily assumes quantisation without explaining it (later explained by de Broglie).
Marking-scheme points
- ✓Works only for hydrogen/single-electron atoms
- ✓Cannot explain fine structure or relative intensities of lines
- ✓Cannot explain Zeeman/Stark effects; quantisation assumed arbitrarily
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Nuclei
State Einstein's mass-energy relation. What is the energy equivalent of 1 atomic mass unit (u)?
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Einstein's mass-energy relation is E = m c^2, which states that mass and energy are interconvertible, where c is the speed of light. Using this relation, the energy equivalent of 1 atomic mass unit (1 u = 1.66 x 10^-27 kg) is about 931 MeV (mega electron volt). This relation explains the large amount of energy released in nuclear reactions such as fission and fusion.
E = m c^2; 1 u = 931 MeV
Marking-scheme points
- ✓E = m c^2 (mass and energy are interconvertible)
- ✓1 u is equivalent to about 931 MeV
- ✓Explains energy released in nuclear reactions
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Nuclei
Define mass defect and binding energy of a nucleus.
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The mass defect is the difference between the sum of the masses of the individual protons and neutrons (nucleons) and the actual mass of the nucleus; the actual nuclear mass is always less than the sum. This missing mass (delta m) is converted into energy that binds the nucleons together. The binding energy is the energy equivalent of the mass defect, BE = (delta m) c^2; it is the energy required to break the nucleus into its constituent nucleons.
BE = (delta m) c^2
Marking-scheme points
- ✓Mass defect = (sum of nucleon masses) - (actual nuclear mass)
- ✓This mass is converted into binding energy
- ✓Binding energy = (delta m) c^2
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Nuclei
The mass defect of a helium nucleus is 0.0304 u. Calculate its binding energy. (1 u = 931 MeV)
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The binding energy is the energy equivalent of the mass defect. BE = (mass defect in u) x 931 MeV = 0.0304 x 931 = 28.3 MeV. Thus the binding energy of the helium nucleus is about 28.3 MeV, and the binding energy per nucleon = 28.3/4 = 7.1 MeV.
BE (MeV) = (delta m in u) x 931
Marking-scheme points
- ✓BE = (mass defect) x 931 MeV
- ✓= 0.0304 x 931
- ✓BE = 28.3 MeV (about 7.1 MeV per nucleon)
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Nuclei
Name the three types of radioactive radiations and state their nature.
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The three types of radioactive radiations are: (1) alpha rays, which are helium nuclei (2 protons + 2 neutrons), positively charged and with low penetrating power; (2) beta rays, which are fast-moving electrons, negatively charged and with greater penetrating power than alpha rays; and (3) gamma rays, which are high-energy electromagnetic waves (photons), electrically neutral and with very high penetrating power. In a magnetic field, alpha and beta rays are deflected in opposite directions while gamma rays are undeflected.
Marking-scheme points
- ✓Alpha: helium nuclei, positive, low penetration
- ✓Beta: fast electrons, negative, moderate penetration
- ✓Gamma: high-energy EM waves, neutral, high penetration
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Nuclei
State the radioactive decay law. What fraction of a radioactive sample remains after 3 half-lives?
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The radioactive decay law states that the rate of decay of a radioactive sample is directly proportional to the number of undecayed nuclei present at that instant: N = N0 e^(-lambda t), where lambda is the decay constant. The half-life T is related to it by T = 0.693/lambda. After each half-life, half of the sample remains, so after 3 half-lives the fraction remaining = (1/2)^3 = 1/8 of the original sample.
N = N0 (1/2)^(t/T)
Marking-scheme points
- ✓Decay law: N = N0 e^(-lambda t); half-life T = 0.693/lambda
- ✓Fraction after n half-lives = (1/2)^n
- ✓After 3 half-lives: (1/2)^3 = 1/8 remains
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Nuclei
What is nuclear fission? Give one example.
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Nuclear fission is the process in which a heavy nucleus (such as uranium-235) splits into two lighter nuclei of comparable masses, with the release of a few neutrons and a large amount of energy. For example, when a uranium-235 nucleus captures a slow neutron, it splits into barium and krypton nuclei plus three neutrons and energy. The released neutrons can cause further fissions, leading to a chain reaction, which is used in nuclear reactors and atom bombs.
Marking-scheme points
- ✓Heavy nucleus splits into two lighter nuclei with energy release
- ✓Example: U-235 + neutron -> lighter nuclei + neutrons + energy
- ✓Released neutrons can cause a chain reaction
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Nuclei
What is nuclear fusion? Why does it require very high temperature?
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Nuclear fusion is the process in which two light nuclei (such as isotopes of hydrogen) combine to form a heavier nucleus, with the release of an enormous amount of energy. It is the source of energy of the sun and stars, where hydrogen nuclei fuse to form helium. It requires very high temperature (millions of degrees) because the positively charged nuclei must overcome their strong electrostatic repulsion to come close enough to fuse.
Marking-scheme points
- ✓Two light nuclei combine into a heavier nucleus with energy release
- ✓Source of energy of the sun and stars (hydrogen to helium)
- ✓Needs very high temperature to overcome electrostatic repulsion
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Semiconductor Electronics
Distinguish between intrinsic and extrinsic semiconductors.
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An intrinsic semiconductor is a pure semiconductor (such as pure silicon or germanium) with no added impurity; its conductivity is low and is due to the equal number of electrons and holes generated thermally. An extrinsic semiconductor is one to which a small amount of a suitable impurity has been added (doping); this greatly increases its conductivity. Extrinsic semiconductors are of two types, n-type and p-type.
Marking-scheme points
- ✓Intrinsic: pure semiconductor, low conductivity, equal electrons and holes
- ✓Extrinsic: doped with impurity, higher conductivity
- ✓Extrinsic types: n-type and p-type
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Semiconductor Electronics
How are n-type and p-type semiconductors formed? Name the majority charge carriers in each.
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An n-type semiconductor is formed by doping a pure semiconductor (silicon) with a pentavalent impurity (such as phosphorus or arsenic), which donates free electrons; the majority carriers are electrons and the minority carriers are holes. A p-type semiconductor is formed by doping with a trivalent impurity (such as boron or aluminium), which creates holes; the majority carriers are holes and the minority carriers are electrons. Both types are electrically neutral overall.
Marking-scheme points
- ✓n-type: pentavalent doping (phosphorus); majority carriers = electrons
- ✓p-type: trivalent doping (boron); majority carriers = holes
- ✓Both are electrically neutral overall
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Semiconductor Electronics
What is a depletion region and potential barrier in a p-n junction?
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When a p-n junction is formed, electrons from the n-side diffuse into the p-side and holes from the p-side diffuse into the n-side, and they recombine near the junction. This leaves a region near the junction that has no free charge carriers but has immobile charged ions; this region is called the depletion region (or depletion layer). The immobile ions set up an internal electric field that opposes further diffusion; the potential difference developed across the depletion region is called the potential barrier.
Marking-scheme points
- ✓Depletion region: layer near the junction with no free carriers
- ✓Formed by diffusion and recombination of electrons and holes
- ✓Potential barrier: potential difference across the depletion region
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Semiconductor Electronics
Distinguish between forward biasing and reverse biasing of a p-n junction diode.
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In forward biasing, the p-side is connected to the positive terminal and the n-side to the negative terminal of the battery; this reduces the width of the depletion region and the potential barrier, so a large current flows and the diode conducts. In reverse biasing, the p-side is connected to the negative terminal and the n-side to the positive terminal; this increases the width of the depletion region and the potential barrier, so only a very small (negligible) current flows and the diode does not conduct.
Marking-scheme points
- ✓Forward bias: p to +, n to -; barrier reduced, diode conducts
- ✓Reverse bias: p to -, n to +; barrier increased, negligible current
- ✓Diode acts as a one-way valve for current
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Semiconductor Electronics
Explain how a p-n junction diode works as a half-wave rectifier.
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A rectifier converts alternating current (AC) into direct current (DC). In a half-wave rectifier, a single diode is connected in series with the AC source and a load resistor. During the positive half-cycle of the AC input, the diode is forward biased and conducts, so current flows through the load. During the negative half-cycle, the diode is reverse biased and does not conduct, so no current flows. As a result, output is obtained only during one half of each cycle, giving a pulsating DC. Its efficiency is low because half the input is wasted.
Marking-scheme points
- ✓Rectifier converts AC into DC; uses one diode
- ✓Positive half-cycle: diode forward biased -> conducts
- ✓Negative half-cycle: diode reverse biased -> no output (pulsating DC)
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Semiconductor Electronics
Explain the working of a full-wave rectifier using two diodes.
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A full-wave rectifier converts both halves of the AC input into DC. It uses two diodes with a centre-tapped transformer and a load resistor. During the positive half-cycle, one diode is forward biased and conducts while the other is reverse biased; during the negative half-cycle, the second diode conducts while the first does not. In both half-cycles, the current through the load flows in the same direction, so output is obtained during the whole cycle. This gives a smoother, more efficient pulsating DC than a half-wave rectifier.
Marking-scheme points
- ✓Uses two diodes and a centre-tapped transformer
- ✓Each diode conducts during one half-cycle
- ✓Current through load is in the same direction for both halves (full-wave DC)
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Semiconductor Electronics
What is a Zener diode? State its main use.
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A Zener diode is a special heavily doped p-n junction diode designed to operate in the reverse breakdown region without being damaged. In this region, the voltage across it remains almost constant (equal to its Zener voltage) even when the current through it changes over a wide range. Because of this property, its main use is as a voltage regulator, that is, to provide a constant output voltage to a load in spite of changes in the input voltage or load current.
Marking-scheme points
- ✓Heavily doped diode that works in reverse breakdown safely
- ✓Voltage across it stays constant (Zener voltage)
- ✓Main use: voltage regulator
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Semiconductor Electronics
Write the truth tables of the OR, AND and NOT logic gates for inputs A and B.
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A logic gate is a digital circuit that gives an output based on its inputs following a logical rule (using binary 0 and 1). OR gate (output Y = A + B): Y = 1 if any input is 1; for inputs (0,0),(0,1),(1,0),(1,1) the outputs are 0,1,1,1. AND gate (Y = A.B): Y = 1 only if both inputs are 1; outputs are 0,0,0,1. NOT gate (Y = not A): it has a single input and inverts it, so input 0 gives output 1 and input 1 gives output 0.
OR: Y = A + B; AND: Y = A.B; NOT: Y = not A
Marking-scheme points
- ✓OR (Y = A + B): output 1 if any input is 1 -> 0,1,1,1
- ✓AND (Y = A.B): output 1 only if both inputs 1 -> 0,0,0,1
- ✓NOT: single input, inverts it (0 -> 1, 1 -> 0)
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