Question

The​ half-life of​ plutonium-241 is approximately 13 years. a. How much of a sample weighing 2 g will remain after 70 ​years? b. How much time is necessary for a sample weighing 2 g to decay to 0.1​ g?

Answer 1

Answer: A) 0.0480 grams and B) 56.16 years.

Explanation: Half live is the time in which the amount of radioactive substance remains halve of its initial amount.

The formula we use for solving this type of problem is:

$\frac{N}{N_0}=(\frac{1}{2})^n$

where, $N_0$ is the initial amount and N is the remaining amount of radioactive substance and n is the number of half lives.

$n=T/t_1_/_2$

where, T is the time and $t_1_/_2$ is half life.

A) from given data, $N_0$ = 2 g

T = 70 years

$t_1_/_2$ = 13 years

N= ?

$n=\frac{70years}{13years}$

n = 5.38

$\frac{N}{2g}=(\frac{1}{2})^5^.^3^8$

$\frac{N}{2g}=0.0240$

N = 0.0480 g

So, 0.0480 grams of the substance will be remaining after 70 years.

B) $N_0$ = 2 g

N = 0.1 g

T = ?

Let's first calculate the value of n for this.

$\frac{0.1}{2}=(\frac{1}{2})^n$

$0.05=0.5^n$

Taking log to both sides:

$log0.05=nlog0.5$

$-1.301=n(-0.3010)$

$n=\frac{1.3010}{0.3010}$

n = 4.32

Half life is 13 years, so we can calculate the time as:

$n=T/t_1_/_2$

$T=n*t_1_/_2$

$T=4.32*13years$

T = 56.16 years

So, it will take 56.16 years for the radioactive substance to decay from 2 g to 0.1 g.

Answer 2

Answer:

Explanation:

a ) 70 years = 70/13 = 5.3846 half years

fraction of matter remaining = (1/2)⁵°³⁸⁴⁶ = 0.02393

g of matter remaining = .02393 x 2 = .0479 g

b ) t = 1/λ ln 2/.1

λ is decay contant and t is time required to convert 2 g to .1 g

λ = .693 / 13 = .0533

t = 1 / .0533 ln 20

= 18.76 x 2.995 = 56.2 years.