A substance has half of 3 min. After 6 min, the count rate was observed to be 400. What was its count rate at zero time? 

Toclay

5 years ago

Half life=3 minutes
2'3=2*2*2=8
After 6minutes,
Then 6minutes divided by 3 minutes=2
After that 8/2=4
4*400=1600

Toclay

5 years ago

I don't understand how u input ur N

DominicAfam

1 year ago

t(1/2) =3
T=6
N=400
No=?😭

n=T/(t1/2)
n=6/3
n=2

N=No/2^n
400=No/2²
No=2²×400
No=4×400
No=1600😂

✌️

PrinceJohn489

1 year ago

We know that the substance has a half-life (t1/2) of 3 minutes. This means that every 3 minutes, the count rate (or activity) decreases by half.

We're given that after 6 minutes, the count rate is 400. We need to find the initial count rate at time 0.

Since 6 minutes is equal to two half-lives (3 minutes x 2), we can use the following formula:

Initial count rate = Final count rate x 2^(Number of half-lives)

Plugging in the values, we get:

Initial count rate = 400 x 2^(2)
= 400 x 4
= 1600

Therefore, the count rate at time 0 was 1600.

Abel004

2 years ago

was it 6mins that we were told to look for or zero min