**IBDP Physics SL 2025 -E.3 Radioactive decay ****SL Paper 1 Exam Style Questions **

## IBDP Physics 2025 SL Paper 1 – All Chapters

## Topic: E.3 Radioactive decay SL Paper 1

Nuclear Biding Energy, Mass-Energy Equivalence, Strong Nuclear Force, Alpha/Beta/Gamma Decay, Half-Life

### Question-E.3 Radioactive decay

A radioactive material has a half-life of 6 days. How long will it take for \(75 \%\) of a pure sample of the material to decay?

A. 3 days

B. 6 days

C. 12 days

D. 18 days

**▶️Answer/Explanation**

Ans:C

The half-life of a radioactive material is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is 6 days.

To find out how long it will take for 75% of the material to decay, we can use the concept of half-life.

After one half-life (6 days), 50% of the material remains, and 50% has decayed. After two half-lives (12 days), another 50% of the remaining material will decay, leaving 25% of the original material.

So, it will take 12 days for 75% of the pure sample of the material to decay.

The correct answer is:C. 12 days.

### 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.

What is the time at which the student measures a count rate of 45 counts per second?

A. \(30 \mathrm{~s}\)

B. \(40 \mathrm{~s}\)

C. \(60 \mathrm{~s}\)

D. \(80 \mathrm{~s}\)

**▶️Answer/Explanation**

Ans:C

Background count rate is 30 , so count rate at t=0 is 120 and at t=20 will be 60 .

From this conclusion clearly we can see half life is 20 sec (time required to half of initial value).

$t_{1/2}=20 ~sec$

\( 120\underset{t_{1/2}} \longrightarrow 60\underset{t_{1/2}} \longrightarrow 30 \underset{t_{1/2}}{\longrightarrow} 15 \)

As background count rate will always present , when count rate will be 15 due to background count rate addition of 30 it will become 45.

Total time $3\times t_{1/2}\Rightarrow 3\times 20 = 60 \mathrm{~s}$

A simple model of the hydrogen atom suggests that the electron orbits the proton. What is the force that keeps the electron in orbit?

A. Electrostatic

B. Gravitational

C. Strong nuclear

D. Centripetal

**Answer/Explanation**

## Markscheme

**A**

The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons provides the necessary centripetal force for electrons to revolve around the nucleus. The electrostatic force is defined as attractive or repulsive forces between particles due to their electric charges.

The number of neutrons and the number of protons in a nucleus of an atom of the isotope of uranium \(_{\;{\text{92}}}^{{\text{235}}}{\text{U}}\) are

**Answer/Explanation**

## Markscheme

**B**

Mass number = sum ( protons no+ neutrons no )

Proton no = 92

Mass number = 235

Neutron = 235 -92

= 143

A radio-isotope has an activity of 400 Bq and a half-life of 8 days. After 32 days the activity of the sample is

A. 200 Bq.

B. 100 Bq.

C. 50 Bq.

D. 25 Bq.

**Answer/Explanation**

## Markscheme

**D**

\(A_{0}\)(intial activity)=400Bq

\(t_{\frac{1}{2}}\)=8days

After 32-days\(=4t_{\frac{1}{2}}\)

Product remain after n half life\(=\frac{A_{0}}{2^{n}}\)

\(=\frac{400}{(2)^{4}}\)

\(=\frac{400}{16}\)

=25Bq