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 :
[tex] ln (\frac{N}{N_0}) = -k * t [/tex]
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 [tex] (T_\frac{1}{2} ) [/tex] and decay constant can be related :
[tex] T_\frac{1}{2} = \frac{ln 2 }{k} = \frac{0.693}{k} [/tex]
Following are the steps can be used to determine amount of radioisotope (N) :
1) To find decay constant :
Given : [tex] (T_\frac{1}{2} ) [/tex] = 28 yrs
Decay constant can be calculated using half life by plugging value in half life formula :
[tex] 28 yrs = \frac{0.693}{k} [/tex]
On multiplying both side by k
[tex] 28 yrs * k= \frac{0.693}{k} *k [/tex]
On dividing both side by 28 yrs
[tex] \frac{28 yrs * k}{28 yrs} = \frac{0.693}{28 yrs} [/tex]
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:
[tex] ln (\frac{N}{3.5 } ) = - 0.02475 yrs^-^1 * 70 yrs [/tex]
[tex] ln (\frac{N}{3.5 } ) = - 1.7325 [/tex]
[tex] ln\frac{N}{No} [/tex] can be converted using the formula ( [tex] ln (\frac{a}{b} ) = ln a - ln b [/tex] )
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
