Question

12(a) Explain the following terms: (i) mass defect; (ii) binding energy of a nucleus; (iii) Assuming the wave nature of an electron, what is the effect of decreasing the speed of a photoelectron on its: (a) wavelength? (b) energy?
(b) A particle of mass 4.4×10⁻²³ kg moves with a velocity of 10⁵ ms⁻¹. Calculate its wavelength. (h=6.6×10⁻³⁴Js)
(c) The diagram above shows part of a radioactive series. Use it to answer the following questions. (i) Name a pair of isotopes. (ii) Name the isotopes with which the series starts. (iii) Write down a nuclear equation for two examples of each of: (a) alpha decay; (b) beta decay.

Accepted Answer

(a)(i) Mass defect (Δm) of a nucleus is the difference between the mass of the atom and the sum of the masses of its constituent parts, i.e. the nucleons of which it is composed.
Δm = (Sum of mass of nucleons) - (Mass of nucleus)

(ii) Binding energy: Binding energy is applicable to subatomic particles in atomic nuclei. It is therefore the minimum energy required to separate the nucleus of an atom into its component parts (neutrons and protons).

(iii) Effect of decreasing speed of photoelectron on:
(a) wavelength ⇒ Wavelength increases, as can be seen from the relation λ=h/mv
(β) Energy ⇒ decreases

(b) Given m=4.4×10⁻²³kg, v=10⁵ms⁻¹, h=6.6×10⁻³⁴Js
λ = h/(mv) = 6.6×10⁻³⁴/(4.4×10⁻²³×10⁵) = 1.5×10⁻¹⁶ m

(c)(i) Pair of isotopes (same proton number, different nucleon number):
²³²₉₀Th and ²²⁸₉₀Th OR ²²⁴₈₈Ra and ²²⁸₈₈Ra OR ²¹²₈₄Po and ²¹⁶₈₄Po OR ²¹²₈₂Pb and ²⁰⁸₈₂Pb

(ii) Isotopes with which the series starts: ²³²₉₀Th and ²²⁸₉₀Th

(iii)(a) Alpha decay examples:
²³²₉₀Th → ²²⁸₈₈Ra + ⁴₂He + energy
²¹⁶₈₄Po → ²¹²₈₂Pb + ⁴₂He + energy

(b) Beta decay examples:
²²⁸₈₈Ra → ²²⁸₈₉Ac + ⁰₋₁e + energy
²¹²₈₂Pb → ²¹²₈₃Bi + ⁰₋₁e + energy

WAEC Physics 2016 Theory — Question 12

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