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

t = 1.4 days

Step-by-step 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

The time take is
`t = 1.3964 \ days`

Step-by-step 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`