Bromine-88 is radioactive and has a half life of 16.3 seconds. What percentage of a sample would be left after 1.74 seconds?Round your answer to 2 significant digits.

To solve this problem, we need to use the formula for exponential decay which is:

N = N0 * (1/2)^(t/T)

where: N is the final quantity remaining after time t, N0 is the initial quantity, T is the half-life of the substance.

Given: T = 16.3 seconds (half-life of Bromine-88), t = 1.74 seconds (time elapsed), N0 = 100% (initial quantity, since we're looking for a percentage).

We need to find N, the final quantity remaining after 1.74 seconds.

Substituting the given values into the formula, we get:

N = 100% * (1/2)^(1.74/16.3)

Now, calculate the exponent first:

1.74/16.3 = 0.1067484662576687

Then, calculate the base to the power of the exponent:

(1/2)^0.1067484662576687 = 0.920044414629323

Finally, multiply this result by 100% to find N:

N = 100% * 0.920044414629323 = 92.00%

So, after 1.74 seconds, approximately 92.00% of the Bromine-88 sample would remain.

To solve this problem, we need to use the formula for exponential decay which is:

N = N0 * (1/2)^(t/T)

where: N is the final quantity of the substance N0 is the initial quantity of the substance t is the time that has passed T is the half-life of the substance

In this case, we want to find N/N0, the fraction of the substance that remains after time t. We know that t = 1.74 seconds and T = 16.3 seconds.

So, we plug these values into the formula:

N/N0 = (1/2)^(1.74/16.3)

Calculating this gives us approximately 0.882, or 88.2% when expressed as a percentage.

So, after 1.74 seconds, about 88.2% of the Bromine-88 sample would remain.