Q.The radius of atom is of the order of 1 Å and radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?

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: $-3 = -3b - 1 + 3b + 2 + 2b$ …