Question

question 5 of 5
in a lab, a scientist is running an experiment on a sample of radioactive particles. she observes that each day, half of the current number of particles disappears. if she wants to have 2 particles of the original sample left on the 5th day, what is the minimum number of particles she should start with?
a 8
b 10
c 16
d 32

Explanation:

Step1: Understand the decay pattern

Each day, half of the particles disappear, so we can work backwards from the 5th day. Let \( N_5 = 2 \) (particles on 5th day).

Step2: Find particles on 4th day

To find the number of particles on the 4th day (\( N_4 \)), since half disappear to get to \( N_5 \), we have \( N_5=\frac{N_4}{2} \), so \( N_4 = 2\times N_5=2\times2 = 4 \).

Step3: Find particles on 3rd day

Using the same logic, \( N_4=\frac{N_3}{2} \), so \( N_3 = 2\times N_4=2\times4 = 8 \).

Step4: Find particles on 2nd day

\( N_3=\frac{N_2}{2} \), so \( N_2 = 2\times N_3=2\times8 = 16 \).

Step5: Find initial particles (\( N_0 \))

\( N_2=\frac{N_1}{2} \), \( N_1 = 2\times N_2 = 32 \)? Wait, no, wait. Wait, we need to go from day 5 back to day 1? Wait, no, let's correct. Wait, the 5th day: let's index the days as day 1, day 2, day 3, day 4, day 5.

Wait, actually, if on day \( n \), the number of particles is \( N_n \), then \( N_{n + 1}=\frac{N_n}{2} \). So to go back, \( N_n = 2\times N_{n+1} \).

We know \( N_5 = 2 \).

Then \( N_4=2\times N_5 = 4 \) (day 4)

\( N_3=2\times N_4 = 8 \) (day 3)

\( N_2=2\times N_3 = 16 \) (day 2)

\( N_1=2\times N_2 = 32 \)? Wait, no, that can't be. Wait, maybe I messed up the day indexing. Wait, the problem says "on the 5th day" she wants 2 particles. So let's think:

Let the initial number be \( x \).

Day 1: \( \frac{x}{2} \)

Day 2: \( \frac{x}{2^2} \)

Day 3: \( \frac{x}{2^3} \)

Day 4: \( \frac{x}{2^4} \)

Day 5: \( \frac{x}{2^4} \)? Wait, no. Wait, day 1: after first day (day 1 ends), number is \( x/2 \). Day 2 ends: \( x/4 \). Day 3 ends: \( x/8 \). Day 4 ends: \( x/16 \). Day 5 ends: \( x/32 \)? Wait, that's not matching. Wait, I think I had the day count wrong.

Wait, the problem says "on the 5th day" she wants 2 particles. So let's re - index:

Let the starting day be day 0 (initial), day 1: after 1 day, day 2: after 2 days, ..., day 5: after 5 days.

The formula for radioactive decay (half - life) is \( N(t)=N_0\times(\frac{1}{2})^t \), where \( N(t) \) is the number of particles at time \( t \), \( N_0 \) is initial number, \( t \) is number of days.

We want \( N(5) = 2 \). So \( 2=N_0\times(\frac{1}{2})^5 \)

\( 2=N_0\times\frac{1}{32} \)

\( N_0=2\times32 = 64 \)? Wait, that's not one of the options. Wait, maybe the problem is that "on the 5th day" means the 5th day of observation, so the number of half - lives is 4? Wait, maybe the days are counted as day 1: after 1 day, day 2: after 2 days, ..., day 5: after 4 days? No, that doesn't make sense.

Wait, let's look at the options. The options are 8,10,16,32. Let's try working backwards from the 5th day:

5th day: 2 particles.

4th day: since half disappears to get to 5th day, 4th day has \( 2\times2 = 4 \)

3rd day: \( 4\times2 = 8 \)

2nd day: \( 8\times2 = 16 \)

1st day: \( 16\times2 = 32 \)? No, but 32 is option D. Wait, but maybe the days are day 1: start, day 2: after 1 day, ..., day 5: after 4 days? No, the problem says "on the 5th day" she wants 2 particles. Wait, maybe the initial day is day 1, and on day 5, the number is 2. So from day 1 to day 5, there are 4 half - lives? Wait, no, day 1: \( N_1 \)

day 2: \( N_1/2 \)

day 3: \( N_1/4 \)

day 4: \( N_1/8 \)

day 5: \( N_1/16 \)

Set \( N_1/16 = 2 \), then \( N_1=32 \)? But 32 is option D. But wait, maybe I made a mistake in the number of half - lives. Wait, the problem says "each day, half of the current number of particles disappears". So day 1: start with \( x \), end of day 1: \( x/2 \)

end of day 2: \( x/4 \)

end of day 3: \( x/8 \)

end of day 4: \( x/16 \)

end of day 5: \( x/32 \)

But we want end of day 5 to be 2, so \( x/32 = 2 \), \( x = 64 \), which is not an option. So maybe the problem means that on the 5th day, before the decay, or maybe the days are counted as day 0 (initial), day 1 (after 1 day), ..., day 4 (after 4 days) to get to day 5? No. Wait, the options include 32 (D) and 16 (C), 8 (A), 10 (B). Let's check the options by plugging in:

Option A: 8.

Day 1: 8

Day 2: 4

Day 3: 2

Day 4: 1

Day 5: 0.5. Not 2.

Option B:10

Day 1:10

Day 2:5

Day 3:2.5

Day 4:1.25

Day 5:0.625. No.

Option C:16

Day 1:16

Day 2:8

Day 3:4

Day 4:2

Day 5:1. No.

Option D:32

Day 1:32

Day 2:16

Day 3:8

Day 4:4

Day 5:2. Ah! There we go. So I messed up the day counting. So day 1: 32, day 2:16, day 3:8, day 4:4, day 5:2. So the 5th day has 2 particles. So the initial number is 32.

Wait, but when I did the formula earlier, I thought day 5 is after 5 days, but actually, day 1 is the first day (initial is day 0?), no, the problem says "start with", then day 1: after first day, day 2: after second day, ..., day 5: after fifth day. So initial (start) is day 0, day 1: after 1 day, ..., day 5: after 5 days. But in the option D, starting with 32, after 5 days (day 5) we have 2. Let's check:

\( N(t)=32\times(\frac{1}{2})^t \), where \( t \) is number of days.

For \( t = 5 \), \( N(5)=32\times\frac{1}{32}=1 \). Wait, no. Wait, no, if \( t = 1 \), \( N(1)=32\times\frac{1}{2}=16 \) (day 1 end)

\( t = 2 \), \( N(2)=8 \) (day 2 end)

\( t = 3 \), \( N(3)=4 \) (day 3 end)

\( t = 4 \), \( N(4)=2 \) (day 4 end)

\( t = 5 \), \( N(5)=1 \) (day 5 end)

Wait, now I'm confused. There's a contradiction. But when we start with 32, day 1 (after 1 day) has 16, day 2:8, day 3:4, day 4:2, day 5:1. But the problem says "on the 5th day" she wants 2 particles. So maybe the "5th day" is day 4? No. Wait, maybe the problem's "5th day" is the day when she checks, and the decay happens once per day, so the number of decays is 4 times to get to the 5th day? Wait, let's re - read the problem: "each day, half of the current number of particles disappears. If she wants to have 2 particles of the original sample left on the 5th day, what is the minimum number of particles she should start with?"

Ah! "left on the 5th day" – so after 4 days of decay? Wait, no. Let's model it as:

Let \( x \) be the initial number.

After day 1: \( x/2 \)

After day 2: \( x/4 \)

After day 3: \( x/8 \)

After day 4: \( x/16 \)

After day 5: \( x/32 \)

We want \( x/32 = 2 \implies x = 64 \), which is not an option. But the options have 32, which gives \( x/16 = 2 \) (after day 4). So maybe the problem has a typo, or I misread the day. Wait, the options include 32, and when we start with 32, after 4 days (day 4) we have 2, but the problem says day 5. But among the options, 32 is the only one that when you start with it, after 4 days (which would be day 4) you have 2, but maybe the problem meant day 4? Or maybe my day counting is wrong.

Wait, let's check the options again. If we start with 32:

Day 1 (first day): 32 (start), then half disappears, so end of day 1:16

Day 2: end of day 2:8

Day 3: end of day 3:4

Day 4: end of day 4:2

Day 5: end of day 5:1

No, that's not 2. Wait, maybe the "5th day" is the day before the decay? Like, on day 5, before the decay happens, she has 2 particles? No, the problem says "half of the current number of particles disappears" each day, so the number of particles at the start of the day is the current number, then half disappears. So start of day 1: \( x \), end of day 1: \( x/2 \)

start of day 2: \( x/2 \), end of day 2: \( x/4 \)

...

start of day 5: \( x/16 \), end of day 5: \( x/32 \)

We want end of day 5:2, so \( x = 64 \). But 64 is not an option. So there must be a mistake in my understanding.

Wait, maybe the problem is that "half of the current number disappears" means that the remaining is half, so it's a half - life, but maybe the number of days is 4? Let's try \( t = 4 \):

\( N(4)=N_0\times(\frac{1}{2})^4 \)

\( 2 = N_0\times\frac{1}{16} \implies N_0 = 32 \). Ah! So maybe the problem meant the 4th day, but it's written as 5th day. Since 32 is an option (D), and that's the result for \( t = 4 \), I think that's the intended problem. So the answer is D.32.

Answer:

D. 32