Radioisotopes are used to diagnoa various illnesses. Iodine-131 is adminsitred to diagnoaw thyriod gland activity. Original dosage contains 280 MBq of Iodine-131. If. One is lost from body then after 6h, there r 274 MBq. What is half life of I-131 to the nearest day.

A = Ao (1/2)^(t/h)

274 = 280 (1/2)^(6/h)
(274/280) = (1/2)^(6/h)

This can now be solved with logarithms, but if you haven't learned those yet, trial and error is the way to go.

To solve this problem, we can use the formula for radioactive decay:

A(t) = A₀ * (1/2)^(t / T₁/₂),

where:
A(t) is the remaining activity after time t,
A₀ is the initial activity,
T₁/₂ is the half-life of the radioactive substance.

Given the initial activity A₀ = 280 MBq and the activity after 6 hours A(6) = 274 MBq, we can solve for the T₁/₂.

Using the formula:

274 = 280 * (1/2)^(6 / T₁/₂).

Now, let's solve for T₁/₂:

(1/2)^(6 / T₁/₂) = 274 / 280,
(1/2)^(6 / T₁/₂) = 0.978571.

To find the value of T₁/₂, we need to take the logarithm of both sides of the equation:

log2((1/2)^(6 / T₁/₂)) = log2(0.978571),
(6 / T₁/₂) * log2(1/2) = log2(0.978571).

We know that log2(1/2) is -1, so the equation becomes:

-6 / T₁/₂ = log2(0.978571).

Simplifying further:

T₁/₂ / 6 = -1 / log2(0.978571),
T₁/₂ = 6 / (-1 / log2(0.978571)).

Now, let's calculate the value of T₁/₂:

T₁/₂ = 6 / (-1 / log2(0.978571)),
T₁/₂ ≈ 6 / (-1 / -0.0293),
T₁/₂ ≈ 6 / 34.07,
T₁/₂ ≈ 0.1761 days.

Therefore, the half-life of Iodine-131 is approximately 0.1761 days, which can be rounded to the nearest day as 0 days (since it is less than half a day).

To find the half-life of iodine-131 (I-131), we can use the formula for radioactive decay:

N(t) = N₀ * (1/2)^(t/h)

Where:
N(t) is the amount of radioactive substance remaining after time t
N₀ is the initial amount of radioactive substance
t is the elapsed time
h is the half-life of the substance

In this case, we are given that the original dosage of I-131 contains 280 MBq and after 6 hours there are 274 MBq remaining. We can use these values to solve for the half-life.

274 = 280 * (1/2)^(6/h)

To simplify the equation, divide both sides by 280:

274/280 = (1/2)^(6/h)

0.97857 = (1/2)^(6/h)

Now, we can take the logarithm of both sides of the equation to get rid of the exponent:

log(0.97857) = log((1/2)^(6/h))

Using the property of logarithms, we can bring down the exponent:

log(0.97857) = (6/h) * log(1/2)

Now, we can solve for the half-life by rearranging the equation:

h = (6 * log(1/2)) / log(0.97857)

Using a calculator, we can evaluate this equation to find the value of h, which represents the half-life of I-131:

h ≈ 8.02 hours

Converting this to the nearest day, we have:

h ≈ 8.02 hours ≈ 8 hours

Therefore, the half-life of I-131 is approximately 8 hours or 1 day to the nearest day.