Q.The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: $1\ \text{Å} = 10^{-10}\ \text{m}$. The size of a hydrogen atom is about $0.5\ \text{Å}$. What is the total atomic volume in $\text{m}^3$ of a mole of hydrogen atoms?

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: $-3 = -3b - 1 + 3b + 2 + 2b$ …