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

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it

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