3. Sodium-24 has a half-life of 15 hours. If a sample of sodium-24 has an original activity of 800 Bq, what will
its activity be after:
i) 15 hours?
ii) 30 hours?
iii) 45 hours?
iv) 60 hours?

See explanation

Explanation:

From the formula;

0.693/t1/2 = 2.303/t log (Ao/A)

t1/2 = half life of Sodium-24

Ao = initial activity of Sodium-24

A= activity of Sodium-24 at time = t

So,

0.693/15 = 2.303/15 log (800/A)

0.0462 = 0.1535  log (800/A)

0.0462/0.1535 =  log (800/A)

0.3 = log (800/A)

Antilog(0.3) =  (800/A)

1.995 =  (800/A)

A = 800/1.995

A = 401 Bq

ii) 0.693/15 = 2.303/30 log (800/A)

0.0462 = 0.0768 log (800/A)

0.0462/0.0768 =  log (800/A)

0.6 =  log (800/A)

Antilog (0.6) =  (800/A)

3.98 =  (800/A)

A = 800/3.98

A = 201 Bq

iii)

0.693/15 = 2.303/45 log (800/A)

0.0462 = 0.0512  log (800/A)

0.0462/0.0512  =  log (800/A)

0.9 = log (800/A)

Antilog (0.9) =  (800/A)

7.94 = (800/A)

A = 800/7.94

A= 100.8 Bq

iv)

0.693/15 = 2.303/60 log (800/A)

0.0462 = 0.038 log (800/A)

0.0462/0.038 = log (800/A)

1.216 = log (800/A)

Antilog(1.216) = (800/A)

16.44 = (800/A)

A = 800/16.44

A = 48.66 Bq

answer: option c.

explanation:

f=ma newton

f=100x2.5=250n