Before You Begin

Question #44796

Difficulty: 8/10 | 🟢 Free

The Question

A pure sample of a radioactive isotope $^{256}\ce{X}$ initially contains $N_0$ nuclei. It undergoes spontaneous symmetric fission to produce two identical nuclei of $^{128}\ce{Y}$ per decay event, with a decay constant $\lambda_1 = \ln 2 \text{ hr}^{-1}$ . The nuclide $^{128}\ce{Y}$ is also radioactive and subsequently decays to a stable nuclide $^{128}\ce{Z}$ via $\beta^-$ emission with a decay constant $\lambda_2 = 2 \ln 2 \text{ hr}^{-1}$ .

Let $t_{max}$ be the time at which the number of $^{128}\ce{Y}$ nuclei reaches its maximum value. Which of the following represents the correct ratio of the number of nuclei $N_X : N_Y : N_Z$ at $t = t_{max}$ ?

Options

A.
$1 : 1 : 1$
B.
$2 : 1 : 1$
C.
$1 : 1 : 0$
D.
$2 : 1 : 3$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Stoichiometric Coefficient Division|Radioactive Decay as Kinetics

Why This Question Is Tricky

Common Mistakes

This problem tests two critical exceptions to standard sequential decay ( $A \rightarrow B \rightarrow C$ ).

First, the stoichiometric multiplier in the rate equation: students often blindly apply the standard Bateman equation $N_Y = \frac{\lambda_1 N_0}{\lambda_2 - \lambda_1}(e^{-\lambda_1 t} - e^{-\lambda_2 t})$ , missing the factor of 2 arising from symmetric fission ( $X \rightarrow 2Y$ ).

Second, the nuclear mass balance: students are habituated to the conservation law $N_X + N_Y + N_Z = N_0$ .

However, due to fission, conservation of mass number dictates $256 N_X + 128 N_Y + 128 N_Z = 256 N_0$ , simplifying to $2N_X + N_Y + N_Z = 2N_0$ .

Failing to account for either the kinetic multiplier or the modified mass balance leads to the distractor options.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Kinetics - Class 12
Concept	Stoichiometric Coefficient Division
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free