Q4. Symbols 7935Br and 79Br can be written, but 3579Br and 35Br are not acceptable. Explain briefly (NCERT Exercise 2.41, page 71).
The left superscript is the mass number and the left subscript the atomic number. For bromine A = 79 and Z = 35, so 7935Br is right. 79Br is also fine, since the symbol Br already fixes Z = 35. 3579Br swaps the two numbers, and 35Br would specify 35 protons and zero neutrons, not atomic number 35 as intended. No bound bromine nucleus with that composition exists; use the mass number 79 for this isotope.
Q5. If the diameter of a carbon atom is 0.15 nm, how many carbon atoms can be placed side by side in a straight line along a scale 20 cm long (NCERT Exercise 2.35, page 71)?
20 cm = 0.20 m and 0.15 nm = 1.5 × 10 m, so the number is (0.20 m) ÷ (1.5 × 10 m) ≈ 1.33 × 10 atoms, or 1.3 × 10 to the two significant figures of 0.15 nm.
Q6. The diameter of a zinc atom is 2.6 Å. Find (a) its radius in pm and (b) the number of atoms in a line 1.6 cm long (NCERT Exercise 2.37, page 71).
(a) 2.6 Å = 260 pm, so the radius is 130 pm. (b) (1.6 × 10 m) ÷ (2.6 × 10 m) ≈ 6.15 × 10 atoms, or 6.2 × 10 to two significant figures.
Q7. An atom contains 29 electrons and 35 neutrons. Find the number of protons and identify the atom (adapted from NCERT Exercise 2.26, page 70).
A neutral atom has as many protons as electrons: 29. Z = 29 is copper, and A = 29 + 35 = 64: 6429Cu.