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

Step-by-step 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.

Answer 2

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:

`(N)/(N_0)=((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=(70years)/(13years)`

n = 5.38

`(N)/(2g)=((1)/(2))^5^.^3^8`

`(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.

`(0.1)/(2)=((1)/(2))^n`

`0.05=0.5^n`

Taking log to both sides:

`log0.05=nlog0.5`

`-1.301=n(-0.3010)`

`n=(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.