MHT-CET Physics · Structure of Atoms and Nuclei

Nuclei: Radioactive Decay and Binding Energy

A radioactive sample loses a fixed fraction of its nuclei per unit time, N = N₀e^(−λt), so it halves every T½ = ln 2/λ; each alpha decay removes 4 from the mass number and 2 from the atomic number, each beta-minus adds 1 to the atomic number, and a nucleus's binding energy is its mass defect times c².

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Why this matters

18 PYQs, one HARD. Eleven are the decay law — the fraction left after some half-lives, a decay constant from two activities, the time for two samples to reach a given ratio, and activity as λN. Five count the alpha and beta particles between two nuclei or name the particle a decay emits, and two are binding energy and fission. Three cards.

Concept 1 of 3: The Decay Law and Half-Life

Each nucleus has the same chance λ per second of decaying, so the number left falls exponentially: N = N₀e^(−λt). After one half-life T½ = ln 2/λ half remain, after n half-lives (1/2)ⁿ. Activity, the decays per second, is A = λN and falls the same way, so two activities give λ directly: A/A₀ = e^(−λt). Two samples with decay constants λ₁ and λ₂ starting equal have N₁/N₂ = e^(−(λ₁ − λ₂)t). Read the question for 'decayed' or 'left': after 3 half-lives, 12.5% is left and 87.5% has decayed.

Definition

$N = N_0 e^{-\lambda t}$ ; $T_{1/2} = \dfrac{\ln 2}{\lambda}$ ; mean life $\dfrac{1}{\lambda}$ .
After n half-lives: left $\left(\tfrac{1}{2}\right)^n$ , decayed $1 - \left(\tfrac{1}{2}\right)^n$ .
Activity $A = \lambda N = \dfrac{\ln 2}{T}N$ : two samples $\dfrac{A_1}{A_2} = \dfrac{N_1T_2}{N_2T_1}$ .
From two activities: $\dfrac{A}{A_0} = e^{-\lambda t}$ (9000 → 3000 in 2 min ⇒ $\lambda = 0.5\ln 3$ per min).
Two samples from equal numbers: $\dfrac{N_1}{N_2} = e^{-(\lambda_1 - \lambda_2)t}$ ( $7\lambda$ and $\lambda$ , ratio e ⇒ $t = \dfrac{1}{6\lambda}$ ).
$\dfrac{dA}{dt} = -\lambda^2 N \propto T^{-2}$ .

Decay law

N = N_0 e^{-\lambda t}, \qquad T_{1/2} = \frac{\ln 2}{\lambda}, \qquad A = \lambda N

Worked example

A sample's activity falls from 12 000 to 3 000 decays per minute in 6 minutes. Half-life and decay constant?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 2 · Q42Moderate

Example 1 · Structure of Atoms and Nuclei · Radioactive Decay and Half-Life

Half-lives of two radioactive elements A and B are 30 minute and 60 minutes respectively. Initially the samples have equal number of nuclei. After 120 minute the ratio of decayed numbers of nuclei of B to that of A will be

Answering 'left' when the question asks 'decayed'

(1/2)ⁿ is what REMAINS. The decayed fraction is 1 − (1/2)ⁿ, and the options carry both.

Multiplying N by T for activity

A = λN = N ln 2/T: a LONGER half-life means a LOWER activity. The ratio of activities is N₁T₂ : N₂T₁, not N₁T₁ : N₂T₂.

Drill 10 more on the decay law and half-life

Concept 2 of 3: Counting Alpha and Beta Decays

Only alpha decay changes the mass number, by 4 each time, so the number of alphas is the fall in mass number divided by 4. Each alpha also lowers the atomic number by 2, and each beta-minus raises it by 1, so the betas make up the difference: β = 2α − (Z_initial − Z_final). Gamma emission changes neither number. Beta-minus emits an electron and an antineutrino; beta-plus emits a positron and a neutrino. One alpha followed by two beta-minus returns the atomic number to where it started, so the first and last nuclei are isotopes.

Definition

α: $A - 4$ , $Z - 2$ . β⁻: $Z + 1$ (electron + antineutrino). β⁺: $Z - 1$ (positron + neutrino). γ: no change.
$n_\alpha = \dfrac{A_i - A_f}{4}$ ; $n_\beta = 2n_\alpha - (Z_i - Z_f)$ .
²³⁸U → ²⁰⁶Pb: 8α, 6β. ²²⁶Ra → ²⁰⁶Pb: 5α, 4β.
Same Z, different A: isotopes; same A: isobars.

Counting decays

n_\alpha = \frac{A_i - A_f}{4}, \qquad n_\beta = 2n_\alpha - (Z_i - Z_f)

Worked example

²³⁵₉₂U decays to ²⁰⁷₈₂Pb. How many alpha and beta particles?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q30Moderate

Example 2 · Structure of Atoms and Nuclei · Radioactive Decay and Half-Life

In the uranium radioactive series, the initial nucleus is

_{92}^{238}\text{U}

and the final nucleus is

_{82}^{206}\text{Pb}

. The number of

\alpha

-particles and

\beta

-particles emitted are

Pairing beta-plus with an antineutrino

Beta-minus (Z rises) comes with an ANTIneutrino; beta-plus (Z falls, a positron) comes with a neutrino.

Counting betas before alphas

Fix the alphas from the mass number first; only then does the change in Z tell you the betas. Starting from Z gives a wrong count.

Drill 4 more on counting alpha and beta decays

Concept 3 of 3: Binding Energy and a Nucleus That Splits

A nucleus weighs less than its separate protons and neutrons. The shortfall, the mass defect Δm = Zm_p + (A − Z)m_n − M, is the binding energy divided by c², with 1 u equal to 931.5 MeV. When a nucleus at rest splits in two, the pieces fly apart with equal and opposite momenta, so their speeds are in the inverse ratio of their masses. Nuclear mass is proportional to A, and the radius to A^(1/3), so a radius ratio of 1 : 2 means a mass ratio of 1 : 8 and speeds of 8 : 1.

Definition

$\Delta m = Zm_p + (A - Z)m_n - M$ ; $BE = \Delta m\,c^2$ , $1$ u $= 931.5$ MeV.
¹⁷₈O: 8 protons, 9 neutrons: $BE = (8M_P + 9M_N - M_O)c^2$ .
$R = R_0A^{1/3}$ , so mass ratio = (radius ratio)³.
Splitting at rest: $m_1v_1 = m_2v_2$ , $\dfrac{v_1}{v_2} = \dfrac{m_2}{m_1}$ .

Binding energy

BE = \left[Zm_p + (A - Z)m_n - M\right]c^2

Worked example

The mass defect of ⁴He is 0.0304 u. Binding energy?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q30Moderate

Example 3 · Structure of Atoms and Nuclei · Radioactive Decay and Half-Life

A nucleus at rest splits into two nuclear parts having radii in the ratio 1:2. Their velocities are in the ratio

Taking the radius ratio as the mass ratio

Mass goes as A and radius as A^(1/3), so the mass ratio is the radius ratio CUBED. Radii 1 : 2 give masses 1 : 8.

Writing the binding energy with the nucleus first

Binding energy is positive: nucleons minus nucleus. One paper prints (M_O − 8M_P − 9M_N)c², which is negative; the magnitude is what is meant.

Drill 1 more on binding energy and a nucleus that splits

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The Decay Law and Half-Life
Decay law
$N = N_0 e^{-\lambda t}, \qquad T_{1/2} = \frac{\ln 2}{\lambda}, \qquad A = \lambda N$
Counting Alpha and Beta Decays
Counting decays
$n_\alpha = \frac{A_i - A_f}{4}, \qquad n_\beta = 2n_\alpha - (Z_i - Z_f)$
Binding Energy and a Nucleus That Splits
Binding energy
$BE = \left[Zm_p + (A - Z)m_n - M\right]c^2$

Watch out for (6)

Answering 'left' when the question asks 'decayed'
→ The Decay Law and Half-Life
Multiplying N by T for activity
→ The Decay Law and Half-Life
Pairing beta-plus with an antineutrino
→ Counting Alpha and Beta Decays
Counting betas before alphas
→ Counting Alpha and Beta Decays
Taking the radius ratio as the mass ratio
→ Binding Energy and a Nucleus That Splits
Writing the binding energy with the nucleus first
→ Binding Energy and a Nucleus That Splits

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Related notes

Bohr Model — the atom around the nucleus →