Home / IBDP Physics- E.3 Radioactive decay- IB Style Questions For HL Paper 1A

IBDP Physics- E.3 Radioactive decay- IB Style Questions For HL Paper 1A -FA 2025

IBDP Physics HL Paper 1 -FA 2025- All Chapters

Question

Three radioactive decay products are:
I. alpha particles, II. beta particles, III. gamma photons.
Which can be deflected by both magnetic and electric fields?
(A) I and II only
(B) I and III only
(C) II and III only
(D) I, II and III
▶️ Answer/Explanation
Detailed solution

Deflection in electric/magnetic fields requires charged particles:
• Alpha particles: positively charged (deflected)
• Beta particles: negatively charged (deflected)
• Gamma photons: neutral (not deflected)
✅ Answer: (A) I and II only

Question

Radioactive nuclide \(X\) decays into a stable nuclide \(Y\). The decay constant of \(X\) is \(\lambda\). The variation with time \(t\) of the number of nuclei of \(X\) and \(Y\) are shown on the same axes.
Radioactive decay graph for X and Y
What is the expression for \(s\)?
(A) \(\dfrac{\ln 2}{\lambda}\)
(B) \(\dfrac{1}{\lambda}\)
(C) \(\dfrac{\lambda}{\ln 2}\)
(D) \(\ln 2\)
▶️ Answer / Explanation
Detailed solution

From the graph, the quantity \(s\) represents the half-life of the radioactive nuclide \(X\), which is the time taken for the number of undecayed nuclei to fall to half its initial value.

For radioactive decay, the relationship between the half-life \(s\) and the decay constant \(\lambda\) is:

\[ s = \frac{\ln 2}{\lambda} \]

This expression follows directly from the exponential decay law \(N = N_0 e^{-\lambda t}\).

✅ Answer: (A)

Question

A student measures the count rate of a radioactive sample with time in a laboratory. The background count in the laboratory is 30 counts per second.
Count rate / counts s−1Time / s
1500
9020
What is the time at which the student measures a count rate of 45 counts per second?
A. \(30\ \text{s}\)
B. \(40\ \text{s}\)
C. \(60\ \text{s}\)
D. \(80\ \text{s}\)
▶️ Answer / Explanation
✅ Answer: C
The measured count rate includes the background count rate of \(30\ \text{counts s}^{-1}\).
Corrected count rate at \(t = 0\): \[ 150 – 30 = 120\ \text{counts s}^{-1} \]
Corrected count rate at \(t = 20\ \text{s}\): \[ 90 – 30 = 60\ \text{counts s}^{-1} \]
The count rate halves from \(120\) to \(60\) in \(20\ \text{s}\), so the half-life is \[ t_{1/2} = 20\ \text{s} \]
Successive halvings: \[ 120 \rightarrow 60 \rightarrow 30 \rightarrow 15 \]
When the corrected count rate is \(15\ \text{counts s}^{-1}\), the measured count rate is \[ 15 + 30 = 45\ \text{counts s}^{-1} \]
This occurs after three half-lives: \[ t = 3 \times 20 = 60\ \text{s} \]
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