CBSE Class 11 Chemistry Structure of Atom Worksheet Set 08

Structure of Atom Worksheet

Teacher's Note:
1. The proton and neutron are each about 1840 times heavier than the electron.
2. In an electric field the neutron is not deflected at all.

Question 2. Describe Rutherford's α-particle scattering experiment. State the observations and the conclusions drawn from them.
Answer:
Experiment. A beam of α-particles from a radioactive source was directed at a very thin gold foil. A circular zinc sulphide screen behind and around the foil showed flashes where the particles landed.
Observations. (i) Most α-particles passed through undeflected. (ii) A small fraction were deflected through small angles. (iii) Very few (about 1 in 20,000) were deflected by large angles, and some rebounded almost straight back.
Conclusions. (i) Most of the atom is empty space. (ii) The positive charge and nearly all the mass are concentrated in a very small region at the centre, the nucleus. (iii) The electrons move round the nucleus, held by electrostatic attraction. (iv) The nuclear radius (about 10⁻¹⁵ m) is about 10⁵ times smaller than the atomic radius (about 10⁻¹⁰ m).
In simple words: A tiny, heavy, positive nucleus sits in a mostly empty atom.

Teacher's Note:
1. Link each observation with its conclusion in the answer.
2. The ratio of the radii (10⁵) is a common one-mark question.

Question 3. State two drawbacks of Rutherford's nuclear model of the atom. Why was Thomson's model rejected?
Answer:
Rutherford's model. (i) An electron revolving round the nucleus is accelerated. By Maxwell's theory an accelerated charge should radiate energy continuously, so the electron should spiral into the nucleus and the atom would collapse. Atoms are stable. (ii) The model does not say how the electrons are arranged around the nucleus or what their energies are.
Thomson's model. It pictured a positive sphere with electrons embedded in it, which could not explain the large-angle scattering of α-particles in Rutherford's experiment, which needs a small, dense nucleus.
In simple words: Rutherford's atom should collapse, and Thomson's atom cannot scatter α-particles through large angles.

Teacher's Note:
1. Stability and electron arrangement are the two standard drawbacks.
2. Bohr's model removed the stability problem by introducing stationary orbits.

Question 4. Find the number of protons, neutrons and electrons in (a) ²⁷Al³⁺ (Z = 13), (b) ³⁵Cl⁻ (Z = 17) and (c) ⁵⁶Fe²⁺ (Z = 26).
Answer:
(a) Protons = 13; neutrons = 27 − 13 = 14; electrons = 13 − 3 = 10.
(b) Protons = 17; neutrons = 35 − 17 = 18; electrons = 17 + 1 = 18.
(c) Protons = 26; neutrons = 56 − 26 = 30; electrons = 26 − 2 = 24.
In simple words: Protons come from Z, neutrons from A − Z, and electrons from Z minus the charge.

Teacher's Note:
1. A positive charge removes electrons; a negative charge adds them.
2. The nucleus is unchanged when an ion forms.

Question 5. Differentiate between isotopes and isobars with one example each. Chlorine has two isotopes, ³⁵Cl (75%) and ³⁷Cl (25%). Calculate its average atomic mass.
Answer:
Isotopes are atoms of the same element with the same atomic number but different mass numbers (different numbers of neutrons), e.g. ¹²C and ¹⁴C. They have the same chemical properties.
Isobars are atoms of different elements with the same mass number but different atomic numbers, e.g. ⁴⁰Ar and ⁴⁰Ca.
Average atomic mass of Cl = (75 × 35 + 25 × 37) ÷ 100 = 35.5 u.
In simple words: Isotopes share Z; isobars share A.

Teacher's Note:
1. Fractional atomic masses arise from the natural mixture of isotopes.
2. The weighted average uses the percentage abundance.

Question 6. An element has mass number 81 and contains 31.7% more neutrons than protons. Assign the atomic symbol.
Answer:
Let the number of protons be x. Neutrons = x + 0.317x = 1.317x.
x + 1.317x = 81, so 2.317x = 81 and x = 34.96, which is 35.
Z = 35, so the element is bromine and the symbol is ⁸¹₃₅Br (neutrons = 46).
In simple words: Write protons plus neutrons equal to the mass number and solve for the protons.

Teacher's Note:
1. Round to the nearest whole number of protons.
2. Check: 46 is 31.4% more than 35 (close to the quoted 31.7%).

Question 7. Blue light has a frequency of 6.5 × 10¹⁴ Hz. Calculate its wavelength (in nm) and wave number (c = 3 × 10⁸ m s⁻¹).
Answer:
λ = c/ν = 3 × 10⁸ ÷ 6.5 × 10¹⁴ = 4.62 × 10⁻⁷ m = 462 nm.
Wave number = 1/λ = 2.17 × 10⁶ m⁻¹.
In simple words: Wavelength is speed divided by frequency, and wave number is its inverse.

Teacher's Note:
1. 1 nm = 10⁻⁹ m.
2. Wave number is the number of wavelengths per metre.

Question 8. Calculate the energy of one photon of (i) light of frequency 3 × 10¹⁵ Hz and (ii) light of wavelength 0.50 Å (h = 6.626 × 10⁻³⁴ J s).
Answer:
(i) E = hν = 6.626 × 10⁻³⁴ × 3 × 10¹⁵ = 1.988 × 10⁻¹⁸ J.
(ii) E = hc/λ = 6.626 × 10⁻³⁴ × 3 × 10⁸ ÷ 0.50 × 10⁻¹⁰ = 3.98 × 10⁻¹⁵ J.
In simple words: Use E = hν when frequency is known and E = hc/λ when wavelength is known.

Teacher's Note:
1. Convert 0.50 Å to 0.50 × 10⁻¹⁰ m.
2. Shorter wavelength means a more energetic photon.

Question 9. A 25 watt bulb emits monochromatic yellow light of wavelength 0.57 μm. Calculate the rate of emission of quanta per second.
Answer:
Energy of one photon = hc/λ = 6.626 × 10⁻³⁴ × 3 × 10⁸ ÷ 0.57 × 10⁻⁶ = 3.49 × 10⁻¹⁹ J.
Power = 25 J s⁻¹, so quanta emitted per second = 25 ÷ 3.49 × 10⁻¹⁹ = 7.17 × 10¹⁹ s⁻¹.
In simple words: Divide the energy emitted every second by the energy of a single photon.

Teacher's Note:
1. 1 μm = 10⁻⁶ m.
2. 1 watt is 1 joule per second.

Question 10. State the photoelectric effect. The work function of caesium is 1.9 eV. Calculate (a) the threshold wavelength, (b) the threshold frequency, and (c) the kinetic energy and speed of the electron ejected by light of wavelength 500 nm (1 eV = 1.602 × 10⁻¹⁹ J).
Answer:
Photoelectric effect. Electrons are ejected from a metal surface when light of frequency above a threshold value falls on it.
W₀ = 1.9 × 1.602 × 10⁻¹⁹ = 3.04 × 10⁻¹⁹ J.
(a) λ₀ = hc/W₀ = 653 nm (6.53 × 10⁻⁷ m).
(b) ν₀ = W₀/h = 4.59 × 10¹⁴ s⁻¹.
(c) K.E. = hc/λ − W₀ = 3.98 × 10⁻¹⁹ − 3.04 × 10⁻¹⁹ = 9.32 × 10⁻²⁰ J.
From ½mv² = K.E., v = √(2 K.E./m) = 4.52 × 10⁵ m s⁻¹.
In simple words: Light of 500 nm has more energy than the work function, so the extra energy becomes the electron's kinetic energy.

Teacher's Note:
1. K.E. = photon energy − work function.
2. Use m = 9.109 × 10⁻³¹ kg for the electron.

Question 11. Explain the observations of the photoelectric effect and how Einstein's photon picture accounts for them.
Answer:
(i) Electrons are emitted at once when light strikes the surface, with no time lag.
(ii) The number of electrons emitted is proportional to the intensity (brightness) of the light.
(iii) Below a threshold frequency ν₀ no electrons are emitted whatever the intensity.
(iv) Above ν₀, the kinetic energy of the electrons increases with the frequency of the light.
Explanation. Light consists of photons of energy hν. One photon gives all its energy to one electron. If hν is less than the work function hν₀ the electron cannot escape. If hν is larger, hν = hν₀ + ½mv². More intense light means more photons and so more electrons, but each photon still has the same energy.
In simple words: One photon knocks out one electron; the photon's energy, set by frequency, decides how fast it leaves.

Teacher's Note:
1. The wave theory could not explain the threshold frequency.
2. Photoelectric effect shows the particle nature of light.

Question 12. Distinguish between (a) continuous and line spectra and (b) emission and absorption spectra.
Answer:
(a) A continuous spectrum contains all wavelengths, merging into one another, like the rainbow from white light. A line spectrum has only certain wavelengths, seen as separate bright lines with dark gaps, and is characteristic of each atom.
(b) An emission spectrum is obtained by analysing the radiation emitted by a substance that has absorbed energy (bright lines). An absorption spectrum is obtained when white light passes through a sample; the wavelengths it absorbs appear as dark lines in an otherwise continuous spectrum.
The study of these spectra is called spectroscopy.
In simple words: Continuous means every colour; line means only a few colours; absorption shows dark lines where light is missing.

Teacher's Note:
1. Each element has its own line spectrum, so it works like a fingerprint.
2. Emission and absorption lines of one element occur at the same wavelengths.

Question 13. Yellow light emitted by a sodium lamp has a wavelength of 580 nm. Calculate its frequency and wave number. Which has more energy per photon, this light or red light of wavelength 700 nm?
Answer:
ν = c/λ = 3 × 10⁸ ÷ 580 × 10⁻⁹ = 5.17 × 10¹⁴ Hz.
Wave number = 1/λ = 1.72 × 10⁶ m⁻¹.
Since E = hc/λ, the shorter wavelength has the higher energy. Yellow light (580 nm) has more energy per photon than red light (700 nm).
In simple words: Frequency is c/λ; energy goes up as wavelength goes down.

Teacher's Note:
1. Do the nm to m conversion before dividing.
2. Red is at the low-energy end of the visible spectrum.