### Question

Radioactive nuclide \(X\) decays into a stable nuclide \(Y\). The decay constant of \(X\) is \(\lambda\). The variation with time \(t\) of number of nuclei of \(X\) and \(Y\) are shown on the same axes.

What is the expression for \(s\) ?

A. \(\frac{\ln 2}{\lambda}\)

B. \(\frac{1}{\lambda}\)

C. \(\frac{\lambda}{\ln 2}\)

D. \(\ln 2\)

**▶️Answer/Explanation**

Ans:A

Certainly! In the context of radioactive decay, the parameter \(s\) represents the half-life of a radioactive substance. The half-life is the amount of time it takes for half of a sample of radioactive nuclei to decay into a stable product. It’s a fundamental concept in nuclear physics and is a measure of how quickly a radioactive substance decays.

The half-life (\(s\)) is related to the decay constant (\(\lambda\)) as follows:

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

Where:

- \(s\) is the half-life of the radioactive substance.
- \(\lambda\) is the decay constant of the substance.
- \(\ln 2\) is the natural logarithm of 2.

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

*Question*

Nuclide X can decay by two routes. In Route 1 alpha (α) decay is followed by beta-minus ((β– ) decay. In Route 2 β – decay is followed by α decay. P and R are the intermediate products and Q and S are the final products.

Which statement is correct?

A Q and S are different isotopes of the same element.

B The mass numbers of X and R are the same.

C The atomic numbers of P and R are the same.

D X and R are different isotopes of the same element.

**Answer/Explanation**

Answer – B

Route 1:

\( X\overset{\alpha }{\rightarrow}P\overset{\beta ^{-}}{\rightarrow}Q \)

\( ^{A}_{Z}X\rightarrow ^{A-4}_{Z-2}X{}’\rightarrow ^{A-4}_{Z-1}X{}” \)

Route 2:

\( X\overset{\beta ^{-} }{\rightarrow}R\overset{\alpha }{\rightarrow}\delta \)

\( ^{A}_{Z}X\overset{\beta ^{-}}{\rightarrow}^{A}_{Z-1}X{}”\overset{\alpha }{\rightarrow}^{A-4}_{Z-3}X{}” \)

The graph shows the variation with time *t* of the activity *A* of a radioactive sample. The energy released in each decay is *E*. The shaded area is equal to *S*.

What does the quantity *S* \( \times \) *E* represent?

A. Average energy produced in 2 s.

B. Average power produced in 2 s.

C. Total energy produced in 2 s.

D. Maximum power produced in 2 s.

**Answer/Explanation**

### Markscheme

C

\(_{\;{\text{6}}}^{{\text{11}}}{\text{C}}\) undergoes \({\beta ^ + }\) decay. The products of this decay are the \({\beta ^ + }\) particle, X and Y. What are X and Y?

**Answer/Explanation**

Answer – B

The mass number will be the same but the atomic number will decrease by 1