A heavy nucleus Q of half-life 20 minutes undergoes alpha-decay with probability of 60% and beta-dec

Q: Why can we just take 60% of the total decayed nuclei? Doesn't the rate change over time?

While the overall rate of decay decreases over time as fewer nuclei remain, the *ratio* of $\alpha$-decays to $\beta$-decays is determined by their respective decay constants ($\lambda_\alpha$ and $\lambda_\beta$). Since these constants don't change, the proportion of nuclei going down each path remains a constant 60:40 split throughout the entire process.

Q: Can I calculate the effective half-life for just the $\alpha$-decay?

You could, but it's unnecessary and often confusing. The given half-life of 20 minutes is the *effective* (or total) half-life of the nucleus Q, accounting for both decay paths. It's much simpler to find the total decayed nuclei first and then apply the 60% probability.

Q: What if the probabilities didn't add up to 100%?

In a closed system where a nucleus must decay into something, the sum of all decay probabilities must equal 100%. If they didn't, it would imply there's another hidden decay pathway not mentioned in the problem.

Q: Why did we subtract 125 from 1000? Why not just use 125?

The formula $N = N_0 / 2^n$ gives you the number of nuclei that are *left over* (undecayed). The question asks for the number of nuclei that *did* decay (specifically via $\alpha$-decay). So, you must subtract the remaining nuclei from the initial amount to find the decayed amount.

Q: Is it possible for a single nucleus to undergo both $\alpha$ and $\beta$ decay simultaneously?

No, a single nucleus of Q will undergo *either* an $\alpha$-decay *or* a $\beta$-decay, not both. The 60% and 40% represent the statistical probability for any single nucleus, which translates to the macroscopic fraction of the entire sample.

Question

A heavy nucleus Q of half-life 20 minutes undergoes alpha-decay with probability of 60% and beta-decay with probability of 40%. Initially, the number of Q nuclei is 1000. The number of alpha-decays of Q in the first one hour is

Options:

(A) 50

(B) 75

(C) 350

(D) 525

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have a sample of a radioactive substance, let's call it nucleus Q. You start with exactly $1000$ of these nuclei. The problem tells us that the half-life of Q is $20 	ext{ minutes}$. We are asked to find out what happens over the course of $1 	ext{ hour}$, which is $60 	ext{ minutes}$.

But there's a twist! This nucleus doesn't just decay in one way; it has two parallel pathways. It can undergo $\alpha$-decay with a $60\%$ probability, or it can undergo $\beta$-decay with a $40\%$ probability. Our goal is to find the exact number of $\alpha$-decays that occur in that first hour.

### The Master Equation

First, we need to figure out how many half-lives fit into our $1 	ext{ hour}$ time frame. Since one half-life is $20 	ext{ minutes}$, we can easily calculate the number of half-lives, $n$:

$$n = \frac{t}{T_{1/2}} = \frac{60}{20} = 3$$

So, exactly $3$ half-lives will pass. Now, remember the fundamental law of radioactive decay? The number of nuclei remaining after $n$ half-lives is given by dividing the initial number by $2^n$. Let's substitute our values:

$$N = \frac{N_0}{2^n} = \frac{1000}{2^3}$$

Since $2^3 = 8$, we find that the remaining number of nuclei is:

$$N = \frac{1000}{8} = 125$$

### Final Calculation

If we started with $1000$ nuclei and only $125$ are left, where did the rest go? They decayed! Let's find the total number of decayed nuclei:

$$N_{dec} = 1000 - 125 = 875$$

Now, here is where mistakes happen. Don't just assume all $875$ nuclei underwent $\alpha$-decay. The problem explicitly states that only $60\%$ of the decays are $\alpha$-decays. Because the probabilities are constant, the fraction of nuclei taking each path remains constant over time.

To find the number of $\alpha$-decays, we simply take $60\%$ of the total decayed nuclei:

$$N_\alpha = 0.60 	imes 875$$

**$$N_\alpha = 525$$**

And there we have it! Out of the $875$ nuclei that decayed, exactly $525$ underwent $\alpha$-decay. If you're curious, the remaining $40\%$ (which is $350$ nuclei) underwent $\beta$-decay.

Animated Solution for Physics - Atoms and Nuclei: A heavy nucleus Q of half-life 20 minutes undergoes alpha-decay with probability of 60% and beta-decay with probability of 40%. Initially, the number of Q nuclei is 1000. The number of alpha-decays of Q in the first one hour is

Select Answer:

Visualized Solution

$N_{0} = 1000$

$n = \frac{t}{T _{1/2}}$

$N = \frac{N _{0}}{2 ^{n}}$

$N = 125$

$N_{d ec} = N_{0} - N$

$N_{α} = 0.6 \times N_{d ec}$

$N_{α} = 525$

$N_{β} = 350$

The Sigma Insight: Radioactivity

Analyzing the Setup

The Master Equation

Final Calculation

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Animated Solution for Physics - Atoms and Nuclei: A heavy nucleus Q of half-life 20 minutes undergoes alpha-decay with probability of 60% and beta-decay with probability of 40%. Initially, the number of Q nuclei is 1000. The number of alpha-decays of Q in the first one hour is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N0​=1000

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=T1/2​t​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N=2nN0​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N=125

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ndec​=N0​−N

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Nα​=0.6×Ndec​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Nα​=525

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Nβ​=350