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3 years ago

3. Sodium-24 has a half-life of 15 hours. If a sample of sodium-24 has an

Physics

1 answer:

sattari [20]3 years ago

3 0

Answer:

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

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Explanation:

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$\Delta U=-3.32\times10^{-3}\ J$

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Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

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Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

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Answer:

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v² = u² + 2as

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v = 38.36 m/s

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