Q.The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding $10^7\ \text{K}$, and its outer surface at a temperature of about $6000\ \text{K}$. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun $= 2.0 \times 10^{30}\ \text{kg}$, radius of the Sun $= 7.0 \times 10^{8}\ \text{m}$.

The key idea here is that a nucleus is modeled as a uniform sphere. For any sphere, the ratio of volume to surface area is simply $R/3$ — a clean geometric fact. Once you know the nuclear radius formula, the rest is arithmetic.

The nuclear radius is given by $R = R_0 A^{1/3}$, where $R_0 = 1.2 \times 10^{-15}$ m and $A$ is the mass number. For Al$^{27}$, $A = 27$, so $A^{1/3} = 3$. That gives $R = 1.2 \times 10^{-15} \times 3 = 3.6 \times 10^{-15}$ m.

Now, for a sphere:

• Volume $V = \frac{4}{3}\pi R^3$
• Surface area $S = 4\pi R^2$

The ratio $\frac{V}{S} = \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{R}{3}$.

So the ratio is simply $\frac{3.6 \times 10^{-15}}{3} = 1.2 \times 10^{-15}$ m. …

The Van der Waals equation corrects the ideal gas law for real gas behaviour. The term $\frac{a}{V^2}$ accounts for intermolecular attraction, and $b$ accounts for the finite volume of molecules. Because the equation is dimensionally consistent, we can extract the dimensions of $a$ and $b$ by looking at how they appear.

1. Find the dimensions of $b$.

   In the term $(V - b)$, we subtract $b$ from $V$. Only quantities with the same dimensions can be added or subtracted. So $b$ must have the same dimensions as volume $V$.

   Volume has dimensions $[L^3]$.

   Hence, $[b] = [L^3]$.

2. Find the dimensions of $a$.

   Look at the term $\left(P + \frac{a}{V^2}\right)$. Again, $P$ and $\frac{a}{V^2}$ must have the same dimensions because they are added.

   Pressure $P$ has dimensions $[ML^{-1}T^{-2}]$ (force per unit area).

   So $\left[\frac{a}{V^2}\right] = [ML^{-1}T^{-2}]$.

   Since $[V^2] = [L^6]$, we get $[a] = [ML^{-1}T^{-2}] \times [L^6] = [ML^5T^{-2}]$.

3. Find the dimensions of $RT$.

   From the ideal gas law $PV = nRT$, for one mole ($n=1$) we have $PV = RT$.

   So $[RT] = [P][V] = [ML^{-1}T^{-2}] \times [L^3] = [ML^2T^{-2}]$.

   This is the same as energy (work), which makes sense — $RT$ is energy per mole.

4. Combine to get $\frac{ab}{RT}$.

   Now put the dimensions together:

   $[ab] = [a][b] = [ML^5T^{-2}] \times [L^3] = [ML^8T^{-2}]$. …

The activity of a radioactive substance measures how many nuclei decay per unit time. Because decay is a random process governed by probability, the activity decreases exponentially with a characteristic time scale set by the half-life.

The key insight is that after each half-life period, exactly half of the remaining active nuclei have decayed. So if we want the activity to drop to some small fraction like 1%, we need to count how many half-lives fit into that reduction.

The mathematical relationship is:

where $A(t)$ is the activity at time $t$, $A_0$ is the initial activity, and $T_{1/2}$ is the half-life.

Alternatively, using the decay constant $\lambda = \frac{\ln 2}{T_{1/2}}$:

Let me work through this step by step:

1. Set up the equation for 1% activity We want $A(t) = 0.01 \, A_0$, so:

Dividing both sides by $A_0$:

2. Take the natural logarithm

Since $0.01 = \frac{1}{100} = 10^{-2}$:

3. Express the decay constant in terms of half-life …

The key idea here is dimensional analysis — a powerful tool that lets us express any physical quantity in terms of chosen fundamental units. When we pick $C$ (velocity), $G$ (gravitational constant), and $h$ (Planck’s constant) as base units, we need to find how density $\rho$ (mass per volume) relates to them. The trick is to write density as $[ \rho ] = C^a G^b h^c$ and solve for $a$, $b$, $c$ using the known dimensions of each quantity.

Let’s recall the dimensions in the standard MLT (mass, length, time) system:

• Velocity $C$: $[C] = L T^{-1}$
• Gravitational constant $G$: from Newton’s law $F = G \frac{m_1 m_2}{r^2}$, we get $[G] = M^{-1} L^3 T^{-2}$
• Planck’s constant $h$: from $E = h \nu$, we have $[h] = M L^2 T^{-1}$
• Density $\rho$: $[\rho] = M L^{-3}$

Now we set up the equation:

1. Write the dimensional equation We assume $[\rho] = [C]^a [G]^b [h]^c$. Substituting dimensions:

2. Expand and collect powers Right side becomes:

3. Equate exponents for M, L, T

   For mass: $1 = -b + c$

   For length: $-3 = a + 3b + 2c$

   For time: $0 = -a - 2b - c$

4. Solve the system

   From the mass equation: $c = 1 + b$

   From the time equation: $a = -2b - c = -2b - (1 + b) = -3b - 1$

   Substitute into the length equation:

Simplify: $-3 = -3b - 1 + 3b + 2 + 2b$ …

The key idea here is the definition of impact parameter in Rutherford's scattering experiment. The impact parameter $b$ is the perpendicular distance between the initial velocity vector of the $\alpha$-particle and the centre of the target nucleus. It tells you how "off-centre" the collision is.

For a head-on collision, the $\alpha$-particle is aimed straight at the nucleus. That means the line of the initial velocity passes directly through the centre of the nucleus. The perpendicular distance from that line to the centre is therefore zero.

1. Recall the definition: Impact parameter $b$ = distance of closest approach if the nucleus were not there — it's the miss distance. Mathematically, if the initial velocity is along a line, $b$ is the perpendicular distance from the nucleus centre to that line.

2. Head-on means zero miss distance: When the $\alpha$-particle is fired straight at the nucleus, there is no "miss" — it's coming right at it. So $b = 0$.

3. Consequences: For $b = 0$, the $\alpha$-particle experiences the maximum repulsive force (Coulomb force) and comes to rest momentarily at the distance of closest approach before being repelled back. This is the case that gives the largest scattering angle ($180^\circ$). …