Q.Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom ($\sim 10^{-10}\ \text{m}$).

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The fundamental force responsible for the stability of the nuclei is (A) Gravitational force (B) Electromagnetic force (C) Strong nuclear force (D) Weak nuclear force

   ›Reveal solutionSolution

   The stability of atomic nuclei is due to the strong nuclear force, which overcomes the electrostatic repulsion between protons; the correct answer is (C).

   The key idea here is that atomic nuclei are packed with positively charged protons that naturally repel each other via the electromagnetic force. If only that force were at play, no nucleus larger than a single proton could exist. Something else must glue the nucleus together—and that something is the strong nuclear force, the most powerful of the four fundamental forces, but one that acts only over extremely short distances (about the size of a nucleus).

   Let’s walk through why each option is or isn’t responsible for nuclear stability.

   1. Gravitational force (A) – Gravity is always attractive, but it is by far the weakest of the fundamental forces. For the tiny masses of protons and neutrons, the gravitational attraction between them is about $10^{36}$ times weaker than the electromagnetic repulsion. It simply cannot hold a nucleus together.

   2. Electromagnetic force (B) – This force causes like charges to repel. In a nucleus, every proton repels every other proton. If this were the only force, the nucleus would instantly fly apart. So the electromagnetic force actually destabilizes the nucleus, not stabilizes it.

   3. Strong nuclear force (C) – This is the correct answer. The strong force acts between all nucleons (protons and neutrons) and is attractive at distances around $10^{-15}$ m (a few femtometers). It is about 100 times stronger than the electromagnetic force at those distances, so it easily overcomes proton–proton repulsion. However, it has a very short range—it drops to nearly zero if nucleons are more than about $2.5 \times 10^{-15}$ m apart. This is why only protons and neutrons that are very close together feel it, and why larger nuclei need extra neutrons to provide more strong-force “glue” without adding more repulsion. …

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2. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If $n, l$ represent the principal and azimuthal quantum numbers respectively, the formula used to know the number of radial nodes possible for a given orbital is (A) $(n-l)$ (B) $(n-l+1)$ (C) $(n-l-1)$ (D) $(n-2)$

   ›Reveal solutionSolution

   The number of radial nodes in an orbital is given by $n - l - 1$, so the correct choice is (C).

   The key idea is that radial nodes are points (actually spherical surfaces) where the radial part of the wavefunction is zero, excluding the origin and infinity. They depend on how many times the radial function changes sign as you move outward from the nucleus.

   Why this formula works:

   The principal quantum number $n$ tells you the total number of nodes (angular + radial) minus one. The azimuthal quantum number $l$ tells you the number of angular nodes. Since total nodes = $n - 1$, and angular nodes = $l$, the remaining nodes — the radial ones — are simply the difference: $(n - 1) - l = n - l - 1$.

   Let’s walk through it step by step.

   1. Recall the node rule for any orbital

      For a given orbital with principal quantum number $n$, the total number of nodes (surfaces where the wavefunction is zero) is $n - 1$. This includes both angular nodes (planes or cones) and radial nodes (spherical shells).

   2. Identify the angular nodes

      The azimuthal quantum number $l$ directly gives the number of angular nodes. For example, an s-orbital ($l=0$) has 0 angular nodes; a p-orbital ($l=1$) has 1 angular node; a d-orbital ($l=2$) has 2, and so on.

   3. Subtract to find radial nodes

      Since total nodes = angular nodes + radial nodes, we have:

   $\text{radial nodes} = (\text{total nodes}) - (\text{angular nodes}) = (n - 1) - l = n - l - 1.$

   4. Check with examples …

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3. MHT-CET 2024Set pcm-2024-05-16-E1 markMCQ
   Q.Which from following is a largest size nanomaterial? (A) Water (molecular level) (B) Glucose (molecular level) (C) Virus (D) Bacteria

   ›Reveal solutionSolution

   Bacteria are largest on the listed scale.

   In increasing size: water molecule < glucose molecule < virus < bacteria. Bacteria (µm-scale) are the largest of …

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4. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.Identify the impossible quantum number set for the electron from the following (A) $n=2,\, l=0,\, m=0,\, s=-\dfrac{1}{2}$ (B) $n=2,\, l=1,\, m=0,\, s=\dfrac{1}{2}$ (C) $n=3,\, l=3,\, m=1,\, s=\dfrac{1}{2}$ (D) $n=4,\, l=2,\, m=1,\, s=\dfrac{1}{2}$

   ›Reveal solutionSolution

   The key idea is that the azimuthal quantum number $l$ must satisfy $0 \le l \le n-1$. Option (C) violates this rule because $l=3$ is not allowed for $n=3$. The impossible set is (C).

   The relevant concept is the quantum number constraints for an electron in an atom. The principal quantum number $n$ (positive integer) sets the shell; the azimuthal quantum number $l$ (integer from 0 to $n-1$) defines the subshell; the magnetic quantum number $m$ (integer from $-l$ to $+l$) gives orbital orientation; and the spin quantum number $s$ is always $\pm \frac{1}{2}$. The most common pitfall is forgetting that $l$ cannot equal or exceed $n$. Here, we check each option against these rules.

   1. Check option (A): $n=2$, $l=0$, $m=0$, $s=-\frac{1}{2}$.

      • For $n=2$, allowed $l$ values are 0 and 1. Here $l=0$ is fine.
      • For $l=0$, allowed $m$ is only 0. So $m=0$ is fine.
      • Spin $s=-\frac{1}{2}$ is allowed. This set is possible.

   2. Check option (B): $n=2$, $l=1$, $m=0$, $s=\frac{1}{2}$.

      • For $n=2$, $l=1$ is allowed (since $l \le n-1 = 1$).
      • For $l=1$, $m$ can be $-1, 0, +1$; $m=0$ is fine.
      • Spin $s=\frac{1}{2}$ is allowed. This set is possible.

   3. Check option (C): $n=3$, $l=3$, $m=1$, $s=\frac{1}{2}$.

      • For $n=3$, the maximum allowed $l$ is $n-1 = 2$. But here $l=3$, which violates the rule $0 \le l \le n-1$.
      • Even though $m=1$ would be fine for $l=3$ (since $|m| \le l$), the illegal $l$ makes the entire set impossible. This set is impossible. …

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5. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.The radius of a nucleus of mass number 27 is $R$. Which of the following is true about a nucleus whose radius is $2R$? (A) It is stable in nature (B) Its mass number is 54 (C) It is likely to undergo fission reaction (D) It is likely to undergo fusion reaction

   ›Reveal solutionSolution

   The nuclear radius scales as $R \propto A^{1/3}$, so doubling the radius multiplies the mass number by 8, giving $A = 216$. A nucleus with such a high mass number is unstable and likely to undergo fission. The correct option is (C).

   The key concept here is the empirical nuclear radius formula: the radius of a nucleus is proportional to the cube root of its mass number, $R = R_0 A^{1/3}$, where $R_0$ is a constant (about $1.2 \times 10^{-15}$ m). This arises because nuclear matter has roughly constant density — like a drop of incompressible liquid — so volume $\propto A$, and since volume $\propto R^3$, we get $R \propto A^{1/3}$.

   Now, let’s work through the problem step by step.

   1. Relate the given radii to mass numbers. We are told a nucleus of mass number $A_1 = 27$ has radius $R$. So:

   $R = R_0 (27)^{1/3} = R_0 \cdot 3.$

   For the second nucleus, radius is $2R$. Let its mass number be $A_2$. Then:

   $2R = R_0 (A_2)^{1/3}.$

   2. Substitute the expression for $R$. From the first equation, $R = 3R_0$. Plug into the second:

   $2(3R_0) = R_0 (A_2)^{1/3} \quad \Rightarrow \quad 6R_0 = R_0 (A_2)^{1/3}.$

   Cancel $R_0$ (non-zero):

   $6 = (A_2)^{1/3}.$

   3. Solve for $A_2$. Cube both sides:

   $6^3 = A_2 \quad \Rightarrow \quad A_2 = 216.$

   So the nucleus with radius $2R$ has mass number 216, not 54. This eliminates option (B).

   4. Interpret the stability and reaction type.
      • Nuclei with mass numbers around 216 are far beyond the iron peak (the most stable nuclei are near $A \approx 56$). Such heavy nuclei are unstable and tend to undergo fission — splitting into smaller, more stable fragments — to release energy. …

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6. MHT-CET 2023Set pcm-2023-05-09-E1 markMCQ
   Q.Identify the CORRECT decreasing order of melting point of cluster of sodium atoms depending on size. (A) Cluster of $10^3$ atoms $>$ cluster of $10^4$ atoms $>$ Bulk sodium (B) Bulk sodium $>$ cluster of $10^4$ atoms $>$ Cluster of $10^3$ atoms (C) Cluster of $10^4$ atoms $>$ Cluster of $10^3$ atoms $>$ Bulk sodium (D) Bulk sodium $>$ cluster of $10^3$ atoms $>$ Cluster of $10^4$ atoms

   ›Reveal solutionSolution

   Smaller clusters melt lower: bulk $>$ $10^4$ $>$ $10^3$.

   Melting point decreases as cluster size decreases, so bulk sodium $>$ cluster of $10^4$ $>$ …

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7. TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
   Q.The ratio of the radii of the $2^{\text{nd}}$ orbits of hydrogen atom and the $3^{\text{rd}}$ orbit of $\mathrm{Li}^{2+}$ ion is (A) $3:4$ (B) $4:3$ (C) $4:1$ (D) $3:1$

   ›Reveal solutionSolution

   The radius of an electron orbit in a hydrogen-like atom is proportional to $n^2/Z$. For H ($n=2, Z=1$) and Li²⁺ ($n=3, Z=3$), the ratio is $(2^2/1) : (3^2/3) = 4 : 3$. The correct option is (B).

   The key idea is that the radius of an electron’s orbit in a hydrogen-like atom (one electron, nuclear charge $Ze$) is given by the Bohr model:

   $r_n = \frac{n^2 a_0}{Z}$

   where $a_0$ is the Bohr radius (a constant). So the radius depends on the square of the principal quantum number $n$ and inversely on the nuclear charge $Z$. This means we don’t need to remember the exact value of $a_0$ — we only need the ratio.

   1. Write the formula for each case For hydrogen atom ($Z=1$), the radius of the $2^{\text{nd}}$ orbit ($n=2$) is:

   $r_{\text{H},2} = \frac{2^2 a_0}{1} = 4a_0$

   For the Li²⁺ ion ($Z=3$), the radius of the $3^{\text{rd}}$ orbit ($n=3$) is:

   $r_{\text{Li}^{2+},3} = \frac{3^2 a_0}{3} = \frac{9a_0}{3} = 3a_0$

   2. Take the ratio $$$$

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8. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.Statement I : The force of attraction due to a hollow spherical shell of uniform density on a point mass situated inside it is always positive. Statement II : The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is same just as if the entire mass of the shell is at the center of the shell. Which of the following is correct? (A) Both statement I and statement II are True (B) Statement I is true, but statement II is false (C) Statement II is true, but statement I is false (D) Both statements, I and II are false

   ›Reveal solutionSolution

   Statement I is false because the net gravitational force inside a hollow spherical shell is zero. Statement II is true because for an external point, a hollow spherical shell behaves gravitationally as if all its mass were concentrated at its center. Therefore, option (C) is correct.

   The problem asks us to evaluate two statements regarding the gravitational force exerted by a hollow spherical shell of uniform density on a point mass. These statements relate to fundamental results in gravitation, often referred to as Newton's Shell Theorem. Understanding these results is crucial for solving problems involving extended masses.

   The core idea behind these results is the principle of superposition and the inverse square nature of the gravitational force. When dealing with an extended object like a spherical shell, we imagine it as being composed of many tiny point masses. The total gravitational force on an external point mass is the vector sum of the forces due to all these tiny point masses. Due to the perfect spherical symmetry and uniform density, these vector sums simplify dramatically in two specific cases: when the point mass is inside the shell, and when it is outside.

   1. Analyze Statement I: The force of attraction due to a hollow spherical shell of uniform density on a point mass situated inside it is always positive.

      • Consider a point mass $m$ located inside a hollow spherical shell of mass $M$ and radius $R$. Let the point mass be at a distance $r < R$ from the center of the shell.
      • A remarkable result of gravitational theory is that the net gravitational force exerted by a uniform hollow spherical shell on any point mass inside it is exactly zero. This means that the gravitational field inside such a shell is also zero.
      • This can be understood intuitively by considering the cancellation of forces. If you draw a cone from the point mass to a small area on the shell, and then extend the cone through the point mass to the opposite side of the shell, it will cut out another area. Although the closer area is smaller, it is also closer, and the farther area is larger but farther away. Due to the inverse square law, these two opposing forces exactly cancel each other out. When this is done for all such pairs, the net force on the point mass inside the shell is zero.
      • Since the net force is zero, it cannot be "always positive". A force of zero means there is no net attraction.
      • Therefore, Statement I is false.
      Watch out

      Gravitational force is a vector quantity. While its magnitude is always non-negative, the statement "always positive" implies a non-zero magnitude and a specific direction. For a point inside a hollow shell, the net force is zero, meaning there is no net attraction. …

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9. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.Assertion (A) : Both rhombic and monoclinic Sulphur have $S_8$ molecules. Reason (R) : They have planar structure. The correct option among the following is (A) (A) is true, (R) is true and (R) is the correct explanation for (A) (B) (A) is true, (R) is true but (R) is not the correct explanation for (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true

   ›Reveal solutionSolution

   Both rhombic and monoclinic sulphur are made of $\mathrm{S_8}$ molecules (A is true), but that ring is a puckered crown, not planar (R is false). Option (C).

   The concept: the $\mathrm{S_8}$ crown

   Sulphur's stable molecular unit is a cyclic $\mathrm{S_8}$ ring. Each sulphur atom forms two single $\mathrm{S-S}$ bonds and keeps two lone pairs. The lone-pair repulsion pushes the bond angle down to about

   $\angle\,\mathrm{S\!-\!S\!-\!S} \approx 105^{\circ}$

   A planar regular octagon would demand an internal angle of $135^{\circ}$. Since sulphur insists on $\approx105^{\circ}$, the ring must buckle — it folds into the famous crown (puckered) shape, with the atoms alternating above and below a mean plane.

   Step 1 — Test the Assertion

   Rhombic sulphur ($\alpha$-S, stable below $369~\mathrm{K}$) and monoclinic sulphur ($\beta$-S, stable above it) are packing polymorphs: the same $\mathrm{S_8}$ crowns arranged differently in the crystal lattice. Both therefore contain $\mathrm{S_8}$ molecules. True.

   Step 2 — Test the Reason …

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10. KCET 2021Set B-21 markMCQ
    Q.In a nuclear reactor heavy nuclei is not used as moderators because (A) They will break up (B) Elastic collision of neutrons with heavy nuclei will not slow them down. (C) The net weight of the reactor would be unbearably high (D) Substances with heavy nuclei do not occur in liquid or gaseous state at room temperature.

    ›Reveal solutionSolution

    A moderator must slow down neutrons via elastic collisions, which requires the moderator nuclei to have a mass comparable to the neutron. Heavy nuclei are too massive to absorb enough kinetic energy in a single collision, making them ineffective — so option (B) is correct.

    The key idea here is how a moderator works. In a nuclear reactor, the fission of uranium-235 produces fast neutrons (with energies around 1–2 MeV). These fast neutrons are not very efficient at causing further fission in uranium-235 — they are more likely to be captured without fission, or to escape. To sustain a chain reaction, we need to slow these neutrons down to thermal energies (about 0.025 eV), where the fission cross-section is much larger. That’s the moderator’s job.

    A moderator slows neutrons by elastic collisions. Think of it like billiard balls: when a moving ball hits a stationary one, the lighter the stationary ball, the more speed it takes away from the moving ball. The best energy transfer happens when the two masses are equal. A hydrogen nucleus (a proton) has almost the same mass as a neutron, so a neutron can lose up to 100% of its energy in a single head-on collision. A carbon nucleus (mass 12 u) is heavier, but still light enough to slow neutrons reasonably well. But a heavy nucleus — say, lead (mass 207 u) — is like a bowling ball hitting a wall: the wall barely moves, so the ball keeps almost all its speed.

    Let’s go through the options one by one.

    1. Option (A): "They will break up"

       Heavy nuclei are stable against breakup from neutron collisions at these energies. A neutron colliding with a heavy nucleus does not have enough energy to cause nuclear fission or disintegration — that requires much higher energies or specific isotopes. So this is not the reason.

    2. Option (B): "Elastic collision of neutrons with heavy nuclei will not slow them down"

       This is the correct physics. In an elastic collision, the fraction of kinetic energy transferred from a neutron (mass $m$) to a stationary nucleus (mass $M$) is given by:

    $\frac{\Delta E}{E} = \frac{4 m M}{(m+M)^2} \cos^2 \theta$

    where $\theta$ is the scattering angle in the centre-of-mass frame. The maximum transfer (when $\theta = 0$) is:

    $\frac{\Delta E_{\text{max}}}{E} = \frac{4 m M}{(m+M)^2}$

    For a heavy nucleus, $M \gg m$, so this fraction becomes very small:

    $\frac{4 m M}{(m+M)^2} \approx \frac{4m}{M} \ll 1$

    For example, with a lead nucleus ($M \approx 207$ u), the maximum energy transfer is only about $4/207 \approx 1.9\%$ per collision. It would take hundreds of collisions to slow a neutron down to thermal energies — impractical. In contrast, with hydrogen ($M = 1$ u), the maximum transfer is 100%, and with carbon ($M = 12$ u), it’s about 28%. Heavy nuclei simply cannot slow neutrons efficiently. …

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11. TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
    Q.The graph of $\ln\left(\dfrac{R}{R_0}\right)$ versus $\ln A$ (R = radius of a nucleus and A = mass number) is (A) Straight line (B) Exponential (C) Parabola (D) Ellipse

    ›Reveal solutionSolution

    The relationship between nuclear radius and mass number is $R = R_0 A^{1/3}$. Taking logs gives a linear equation, so the graph is a straight line. The correct option is (A).

    The key idea here is the empirical nuclear radius formula:

    $R = R_0 A^{1/3}$

    where $R_0$ is a constant (about $1.2 \times 10^{-15}$ m). This tells us that the radius grows as the cube root of the mass number. When we take natural logs of both sides, the cube root becomes a simple factor of $\frac{1}{3}$, turning a power law into a straight line. That’s why plotting $\ln(R/R_0)$ against $\ln A$ yields a linear graph — it’s a classic log-log plot of a power function.

    Let’s work through it step by step:

    1. Start with the nuclear radius formula

    $R = R_0 A^{1/3}$

    This is a well-established experimental result: the volume of a nucleus is proportional to its mass number, so the radius scales as $A^{1/3}$.

    2. Divide both sides by $R_0$

    $\frac{R}{R_0} = A^{1/3}$

    This isolates the dimensionless ratio we’re plotting.

    3. Take the natural logarithm of both sides

    $\ln\left(\frac{R}{R_0}\right) = \ln\left(A^{1/3}\right)$

    Using the logarithm power rule: $\ln(x^p) = p \ln x$.

    4. Simplify the right-hand side

    $\ln\left(\frac{R}{R_0}\right) = \frac{1}{3} \ln A$

    This is now in the form $y = m x$, where:

    • $y = \ln(R/R_0)$
    • $x = \ln A$
    • $m = \frac{1}{3}$ (the slope)

    5. Interpret the graph …

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12. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
    Q.Which of the following statement is correct? (A) Electromagnetic force is short ranged (B) Relative strength of gravitational force is higher than that of weak nuclear force (C) Range of the weak nuclear force is smaller than that of strong nuclear force (D) Relative strength of strong nuclear force may or may not be higher than that of electromagnetic force

    ›Reveal solutionSolution

    The key idea is comparing the four fundamental forces by their range and relative strength. The correct statement is that the weak nuclear force has a shorter range than the strong nuclear force, making option (C) the answer.

    The four fundamental forces of nature — gravitational, electromagnetic, strong nuclear, and weak nuclear — differ dramatically in both how far they act and how powerfully they bind. To answer this question, you need a clear mental map of these two properties for each force.

    Range tells you the maximum distance over which the force is effective. Relative strength compares how strong one force is compared to another, usually taking the strong nuclear force as the reference (strength = 1).

    Let’s examine each statement one by one.

    1. Option (A): "Electromagnetic force is short ranged"

       This is false. The electromagnetic force obeys an inverse-square law ($F \propto 1/r^2$) and has infinite range, just like gravity. It can act across atoms, rooms, planets, and galaxies. A short-ranged force dies off extremely rapidly beyond a tiny distance — electromagnetic force does not.

    2. Option (B): "Relative strength of gravitational force is higher than that of weak nuclear force"

       This is false. If we set the strong nuclear force’s strength to 1, the approximate relative strengths are:

       • Strong nuclear: $1$
       • Electromagnetic: $10^{-2}$
       • Weak nuclear: $10^{-5}$
       • Gravitational: $10^{-38}$ Gravity is by far the weakest — about $10^{33}$ times weaker than the weak nuclear force. So gravitational strength is far lower, not higher.

    3. Option (C): "Range of the weak nuclear force is smaller than that of strong nuclear force"

       This is true. The strong nuclear force has a range of about $10^{-15}$ m (roughly the diameter of a medium-sized nucleus). The weak nuclear force has an even shorter range, around $10^{-18}$ m. So the weak force operates over a smaller distance than the strong force. …

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