Q.Which of the following statement is not correct about the characteristics of cathode rays?

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. …

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:

4. Check with examples …

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. …

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:

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

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

Cancel $R_0$ (non-zero):

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

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. …

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:

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:

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

2. Take the ratio $$$$

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. …

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

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 …

The key idea here is the empirical nuclear radius formula:

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

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 $R_0$

This isolates the dimensionless ratio we’re plotting.

3. Take the natural logarithm of both sides

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

4. Simplify the right-hand side

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 …

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. …

The fundamental concept behind this question is mass defect and nuclear binding energy, which are direct consequences of Einstein's mass-energy equivalence principle, $E=mc^2$.

When protons and neutrons (collectively called nucleons) combine to form a stable nucleus, they are held together by the strong nuclear force. This process involves a release of energy, known as the binding energy.

Here's why this leads to a mass difference:

1. Energy Release During Nucleus Formation: Imagine you have a collection of free protons and neutrons. When these individual nucleons come together to form a nucleus, they become bound. For them to be bound, energy must be released from the system. This released energy is the nuclear binding energy.

2. Mass-Energy Equivalence: According to Einstein's famous equation, $E=mc^2$, mass and energy are interchangeable. If energy ($E$) is released when a nucleus forms, it means that a corresponding amount of mass ($m$) must have been converted into that energy.

3. Mass Defect: Therefore, the mass of the newly formed nucleus must be less than the sum of the masses of its individual, unbound constituent protons and neutrons. This difference in mass is called the mass defect ($\Delta m$).

   The mass defect $\Delta m$ for a nucleus with $Z$ protons and $N$ neutrons, and a nuclear mass $M_{nucleus}$, is given by:

   $\Delta m = (Z m_p + N m_n) - M_{nucleus}$

   where $m_p$ is the mass of a proton and $m_n$ is the mass of a neutron.

4. Binding Energy Calculation: The binding energy ($E_b$) of the nucleus is then directly related to this mass defect:

   $E_b = \Delta m c^2$

   This binding energy is the energy required to break the nucleus back into its individual protons and neutrons, or the energy released when the nucleus is formed from its constituent nucleons. …

The Bohr model gives a simple, beautiful scaling law for one-electron ions: the radius of the $n$-th orbit is proportional to $n^2$ and inversely proportional to the nuclear charge $Z$. For hydrogen ($Z=1$, $n=1$), the Bohr radius is $a_0 = 53$ pm. For any other hydrogen-like ion (one electron, nuclear charge $Z$), the Coulomb attraction is stronger, pulling the electron closer.

Be$^{3+}$ is a beryllium atom that has lost three electrons, leaving just one electron orbiting a nucleus with $Z=4$. It is exactly hydrogen-like, so Bohr’s formula applies directly.

1. Recall the Bohr radius formula for a hydrogen-like ion The radius of the $n$-th orbit is

where $a_0 = 53$ pm is the Bohr radius for hydrogen. This comes from balancing centripetal force with Coulomb force and quantising angular momentum — the $n^2$ comes from the energy levels, and the $Z$ in the denominator reflects the stronger pull from a larger nuclear charge.

2. Identify the quantum numbers

   The ground state means $n = 1$. For Be$^{3+}$, the atomic number is $Z = 4$.

3. Substitute and compute

4. Match to the given options …