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.

Explanation:

All the radioactive decay processes follows first order kinetics.

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

$k=\frac{0.693}{t_{1/2}}$

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:

$k=\frac{0.693}{14days}=0.0495days^{-1}$

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

$N=N_o\times e^{-kt}$

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:

$N=124mg\times e^{(-0.0495days^{-1}\times 56days)}=7.75mg$

Hence, the correct answer is Option A.