If 15% of the radioactive material decays in 5 days, what would be the percentage of amount of original material left after 25 days?

ln(No/N) = kt

No = 100
N = 85
t = 5 days
solve for k

Then
ln(No/N) = kt
No = 100
N = ? solve for this amount remaining.
k = from above
t = 25 days

N is what is left. (N/100)*100 = % remaining

how the N becomes 85 in first ?

No clarity

To find the percentage of the original material left after 25 days, we need to understand how the material decays and calculate it step by step.

We are given that 15% of the radioactive material decays in 5 days. This means that after 5 days, 15% of the original material is gone, and 85% of the original material remains.

Now, we will divide the time period of 25 days into smaller intervals that each represent 5 days.

25 days / 5 days = 5 intervals.

In each interval of 5 days, 15% of the remaining material will decay. Since we have 5 intervals, we will have 5 decay cycles.

For each decay cycle, we multiply the remaining material by (1 - 15%) or 0.85.

Let's calculate the percentage of the original material left after 25 days:

Initial amount of material = 100%

First decay cycle (after 5 days): 100% * 0.85 = 85%

Second decay cycle (after 10 days): 85% * 0.85 = 72.25%

Third decay cycle (after 15 days): 72.25% * 0.85 = 61.41%

Fourth decay cycle (after 20 days): 61.41% * 0.85 = 52.2%

Fifth decay cycle (after 25 days): 52.2% * 0.85 = 44.37%

Therefore, after 25 days, approximately 44.37% of the original material would be left.

n(No/N) = kt

No = 100
N = 85
t = 5 days
solve for k

Then
ln(No/N) = kt
No = 100
N = ? solve for this amount remaining.
k = from above
t = 25 days

N is what is left. (N/100)*100 = % left