Question

A 3.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific test site. The half-life of the radioactive element is 28 years. How much of the original sample will remain in the year 2030? Choose the closest.

DO NOT GUESS, ONLY COMMENT IF YOU KNOW

Answer 1

0.50 g

That should be the correct answer :)

Answer 2

Answer : The correct answer for amount of radioisotope remain in 2030 is 0.619 g .

Radioactive Decay is emission of radiations ( in form of alpha , beta particle etc ) by unstable atom .

Radioactive decay is FIRST ORDER reaction . So , the equation of first order can be used to find decay constant , amount of radioisotopes or half life .

The equation for radioactive decay is given as :

`ln ((N)/(N_0)) = -k * t`

Where : N = amount of radioisotope at time t

N₀ = amount of radioisotope initially present

k = decay constant t = time

Half life :

It is time when amount of radioisotope decrease to 50 % of its original amount . Half life
`(T_(1)/(2) )` and decay constant can be related :

`T_(1)/(2) = (ln 2 )/(k) = (0.693)/(k)`

Following are the steps can be used to determine amount of radioisotope (N) :

1) To find decay constant :

Given :
`(T_(1)/(2) )` = 28 yrs

Decay constant can be calculated using half life by plugging value in half life formula :

`28 yrs = (0.693)/(k)`

On multiplying both side by k

`28 yrs * k= (0.693)/(k) *k`

On dividing both side by 28 yrs

`(28 yrs * k)/(28 yrs) = (0.693)/(28 yrs)`

k = 0.02475 yrs⁻¹

2) To find amount of radioisotope (N):

Given : Amount of radioisotope originally present = 3.5 g

Time = 2030 - 1960 = 70 yrs

decay constant = 0.02475 yrs⁻¹

Amount of radioisotope (N) = ?

Plugging these values in the formula as:

`ln ((N)/(3.5 ) ) = - 0.02475 yrs^-^1 * 70 yrs`

`ln ((N)/(3.5 ) ) = - 1.7325`

`ln(N)/(No)` can be converted using the formula (
`ln ((a)/(b) ) = ln a - ln b` )

ln N - ln (3.5 ) = - 1.7325

(ln 3.5 = 1.253 )

ln N -1.253 = -1.7325

Adding both side 1.253

ln N -1.253 + 1.253 = -1.7325 + 1.253

ln N = -0.4795

Taking anti ln of -0.4795

N = 0.619 g

Hence amount of radioisotope remained in 2030 is 0.619 g