Numerical Problems Based on Indirect Methods for Small Distances for Class 11 Physics

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Numerical Problems on Indirect Methods for Small Distances

Class 11 Physics · Units and Measurements
SQ

Atomic Volume and Film Thickness

Apply volume formulas to estimate molecular dimensions
Ch 2 · Units and Measurements
1
The radius of a muonic hydrogen atom is $2.5 \times 10^{-13}\text{ m}$. What is the total atomic volume in $\text{m}^3$ of a mole of such hydrogen atoms?
Answer $3.94 \times 10^{-14}\text{ m}^3$ 📝
Detailed Solution

Given the radius of one atom, $r = 2.5 \times 10^{-13}\text{ m}$.

The volume of a single atom (assuming it to be a sphere) is:

$$V_{\text{atom}} = \frac{4}{3} \pi r^3$$

Number of atoms in 1 mole (Avogadro’s number), $N = 6.023 \times 10^{23}$.

Total volume of 1 mole of H-atoms:

$$V_{\text{total}} = N \times \frac{4}{3} \pi r^3$$
$$V_{\text{total}} = 6.023 \times 10^{23} \times \frac{4}{3} \times 3.14 \times (2.5 \times 10^{-13})^3$$
$$V_{\text{total}} \approx 3.94 \times 10^{-14}\text{ m}^3$$
2
A drop of olive oil of radius $0.25\text{ mm}$ spreads into a circular film of radius $10\text{ cm}$ on the water surface. Estimate the molecular size of olive oil.
Answer $2.08 \times 10^{-7}\text{ cm}$ 📝
Detailed Solution

Assuming that the film has one molecular thickness, the molecular size of the olive oil is equal to the thickness of the film ($t$).

$$t = \frac{\text{Volume of oil drop}}{\text{Area of the circular film}}$$

Radius of drop $r = 0.25\text{ mm} = 0.025\text{ cm}$.
Radius of film $R = 10\text{ cm}$.

$$t = \frac{\frac{4}{3} \pi r^3}{\pi R^2} = \frac{\frac{4}{3} \pi (0.025)^3}{\pi (10)^2}$$
$$t = \frac{4}{3} \times (25 \times 10^{-3})^3 \times 10^{-2}$$
$$t = \frac{4}{3} \times 15625 \times 10^{-9} \times 10^{-2} \approx 2.08 \times 10^{-7}\text{ cm}$$
3
A drop of olive oil of radius $0.30\text{ mm}$ spreads into a rectangular film of $30\text{ cm} \times 15\text{ cm}$ on the water surface. Estimate the molecular size of olive oil.
Answer $2.51 \times 10^{-7}\text{ cm}$ 📝
Detailed Solution

Radius of the oil drop, $r = 0.30\text{ mm} = 0.03\text{ cm}$.

Volume of the single drop ($V$):

$$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.1416 \times (0.03)^3$$
$$V = \frac{4}{3} \times 3.1416 \times (27 \times 10^{-6}) \approx 1.13 \times 10^{-4}\text{ cm}^3$$

Area of the rectangular film ($A$):

$$A = l \times b = 30\text{ cm} \times 15\text{ cm} = 450\text{ cm}^2$$

The molecular size (thickness $t$):

$$t = \frac{V}{A} = \frac{1.13 \times 10^{-4}}{450}$$
$$t \approx 0.00251 \times 10^{-4}\text{ cm} = 2.51 \times 10^{-7}\text{ cm}$$

(Note: The textbook answer key provides $7.5 \times 10^{-7}\text{ cm}$. This is exactly 3 times the calculated value, implying the problem source likely intended for there to be 3 drops of oil, despite the prompt reading “A drop”.)

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