Question

a radioactive sample decays 63% of its initial value in 10 sec , it would have decayed by 50% of its initial value in​

Answer 1

It would have decayed by 50% of its initial value in​ 7 sec.

Explanation:

=> Suppose, the initial amount of radio active sample is x.

According to the question,  it decays 63% of its initial value in 10 sec.

Thus, remaining amount = 100 - 63 = 37

=> No of half lives to obtain 37%:

N = log ( x / 0.35x) / log(2)

   = 1.4344 half lives.

=> it would have decayed by 50% of its initial value in​ time :

1.4344 half lives = 10 s

1 half life = ?

= 1 * 10 / 1.4344

= 6.971

≈ 7 sec

Therefore, A given radioactive sample would have decayed by 50% of its initial value in​ 7 sec.

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Answer 2

The radioactive sample would have decayed by 50% of its initial value in 7 s.

Explanation:

Let initial amount=x

Remaining amount=100-63%=37%

After 10 s

The remaining amount=$x-\frac{63}{100}\times 100=\frac{100x-63x}{100}=0.37 x$

Number of half lives to get 37%

Number of half lives=$\frac{log\frac{Initial\;value}{remaining\;amount}}{\log 2}$

$N=\frac{;og\frac{x}{0.37x}}{log 2}=1.4344 lives$

1.4344 half lives takes time to decay=10 s

1 half life required time to decay=$\frac{10}{1.4344}=6.97 \approx 7s$

Hence, it would have decayed by 50% of its initial value in 7s.

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