Question

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.

Answer 1

Answer:

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

Answer 2

Answer:

The time take is  $t = 1.3964 \ days$

Explanation:

From the question we are told that

    The decay constant is  $\lambda = 0.16$

     The percentage fall is  $c = 0.80$

The equation for radioactive decay is mathematically represented as

               $N(t) = N_o * e^{- \lambda t }$

Where is $N(t)$ is the new amount of the new the isotope while $N_o$ is the original

At initial  $N_o = 100$%  = 1

At $N(t ) = 80$%  = 0.80  

       $0.80 = 1 * e^{- 0.16 t }$

=>     $-0.223 = -0.16 t$

=>     $t = 1.3964 \ days$