Question

A sample of phosphorus-32 has a half-life of 14.28 days. If 55 g of this radioisotope remain unchanged after approximately 57 days, what was the mass of the original sample?

Answer 1

55= No (1/2)^55/57
55= No (1/2)^3.9
55= No (1/2)^4
55= No (1/16)
No= 880 g

Answer 2

Answer: The mass of the original sample is
`12.82* 10^(7) grams`.

Step-by-step explanation:

Half-life of sample of phosphorus-32 = 14.28 days

`\lambda =\frac{0.693}{t_{(1)/(2)}}=(0.693)/(14.28)=0.0485 day^(-1)`

`N=N_o* e^(-\lambda t)`

`lnN=lnN_o-\lambda t`

N = 55 g, t= 57 days

`\ln[55 g]=-0.0485 day^(-1)* 57 days+ln[N_o]`

`\log(55 g)/(N_o)=2.303* (-0.0485 day^(-1))* 57 days`

`(55 g)/(N_o)=antilog[-6.3666]`

`N_o=(55 g)/(4.29* 10^(-7))=12.82* 10^(7) grams`

The mass of the original sample is
`12.82* 10^(7) grams`.