When time is measured in days, the decay constant for a particular radioactive isotope is 0.16. Determine the time required for a confined sample of the isotope to fall to 80% of its original value.

t = 1.4 days

Explanation:

The law of radioactive decay gives the amount of radioactive substance, left after a certain amount of time has passed. The formula of law of radioactive decay is given as follows:

N = N₀ (e)^-λt

where,

λ = decay constant = 0.16

N₀ = Initial Amount of the Substance

N = The Amount of Substance Left after Decay = 80% of N₀ = 0.8 N₀

t = Time Required by the Substance to decay to final value = ?

Substituting these values in the law of radioactive decay formula, we get:

0.8 N₀ = N₀ (e)^-0.16 t

0.8 = (e)^-0.16 t

ln (0.8) = -0.16 t

t = - 0.2231/-0.16

t = 1.4 days

we need some answer choices, or something like that

explanation:

more than, less

less than, more