Question

11. Phosphorus-32 is radioactive and has a half life of 14 days. How much of a 124 mg sample of phosphorus-32 is present after 56 days? A) 7.75 mg B) 15.5 mg C) 31.0 mg D) 62.0 mg

Answer 1

Answer: The correct answer is Option A.

Step-by-step explanation:

All the radioactive decay processes follows first order kinetics.

To calculate the rate constant for a reaction, we use the equation:

`k=(0.693)/(t_(1/2))`

where,

k = rate constant for a reaction

`t_(1/2)` = half life of a reaction = 14 days

Putting all the values in above equation, we get:

`k=(0.693)/(14days)=0.0495days^(-1)`

To calculate the amount of sample left, we use the equation:

`N=N_o* e^(-kt)`

where,

N = amount of sample left after time 't'

`N_o` = initial amount of the sample = 124 mg

k = rate constant of the reaction =
`0.0495days^(-1)`

t = time taken = 56 days

Putting values in above equation, we get:

`N=124mg* e^{(-0.0495days^(-1)* 56days)}=7.75mg`

Hence, the correct answer is Option A.