Logo
Jeeify
NVAEasyJEE 2023Radioactive Decay & Half-Life

JEE Physics 2023 Question with Solution

A radioactive nucleus decays by two different processes. The half-life of the first process is 5 minutes5 \, \text{minutes} and that of the second process is 30 s30 \, \text{s}. The effective half-life of the nucleus is calculated to be α11\frac{\alpha}{11} seconds. The value of α\alpha is _____

Answer

Correct answer:300

Step-by-step solution

Standard Method

Given: The two half-lives are 300 s300 \, \text{s} and 30 s30 \, \text{s}.

Find: The value of α\alpha if the effective half-life is α11\frac{\alpha}{11} seconds.

For two simultaneous decay processes, the effective decay constant is the sum of individual decay constants.

dN1dt=−λ1N\frac{dN_1}{dt} = -\lambda_1 NdN2dt=−λ2N\frac{dN_2}{dt} = -\lambda_2 NdNdt=−(λ1+λ2)N\frac{dN}{dt} = -(\lambda_1 + \lambda_2)N

Hence,

λeq=λ1+λ2\lambda_{eq} = \lambda_1 + \lambda_2

Using the relation between decay constant and half-life as written in the solution,

1t1/2=1300+130=11300\frac{1}{t_{1/2}} = \frac{1}{300} + \frac{1}{30} = \frac{11}{300}

Therefore,

t1/2=30011  st_{1/2} = \frac{300}{11} \; \text{s}

Comparing with α11\frac{\alpha}{11} seconds, we get α=300\alpha = 300.

Therefore, the required numerical value is 300300.

Half-life Addition Shortcut

Given: Two parallel decay modes with half-lives 300 s300 \, \text{s} and 30 s30 \, \text{s}.

Find: α\alpha in effective half-life α11\frac{\alpha}{11}.

For parallel radioactive decay channels, reciprocals of half-lives add in the same way as decay constants:

1teff=1300+130=11300\frac{1}{t_{eff}} = \frac{1}{300} + \frac{1}{30} = \frac{11}{300}

So,

teff=30011t_{eff} = \frac{300}{11}

Thus, α=300\alpha = 300. The required numerical value is 300300.

Common mistakes

  • Adding the half-lives directly as 300+30300 + 30 is incorrect because simultaneous decay processes add through decay constants, not through half-lives. Convert the idea to reciprocal form or use λeq=λ1+λ2\lambda_{eq} = \lambda_1 + \lambda_2.

  • Forgetting to convert 5 minutes5 \, \text{minutes} into 300 s300 \, \text{s} gives a wrong numerical result. Keep all times in the same unit before substitution.

  • Taking the effective half-life as larger than both individual half-lives is conceptually wrong. When two decay processes act together, decay becomes faster, so the effective half-life must be smaller than 30 s30 \, \text{s}? No—the extracted solution uses reciprocal addition of the listed values and yields 30011 s\frac{300}{11} \, \text{s}, which is indeed smaller than 30 s30 \, \text{s}.

Practice more Radioactive Decay & Half-Life questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Radioactive Decay & Half-Life

Related questions