Question

A radioactive sample decays by 63% of its initial value in 10 s. It would have decayed by 50% of its initial value in :
7
O 14
0.7
O 1.4​

Answer 1

$7s$

Explanation:

Given

Initial amount $= x$

after time interval of 10s

Amount  $= 37$

Number of half lives to attain $37\%$

$N = \frac{log( \frac{x}{0.37x})} {log(2)}$

$N = 1.4344$ half lives.

$1.4344$ half lives $= 10s$

$1$ half live $= 6.97s$ (approx. 7)

There is a delay of 50%  in 7 seconds.

Answer 2

A radioactive sample decay by 50% of it's initial value in 7 seconds  

Explanation:

$N(t)=N_o(a)^{t/t_h}$

A radioactive sample decays by 63% of its initial value in 10 s.

$N(t)=37%\ of \ N_o$

$0.37=a^{10/t_h}$

$a^{1/t_h}=0.37^{1/10}\approx 0.9053$

• For same radioactive sample.

It would be decay by 50% in time t

$0.5=(a)^{t/t_{h}}$

$0.5=(a^{1/t_h})^t$

$0.5=(0.9053)^t$

Taking ln both sides

$\ln 0.5=t\ln 0.9053$

$t=6.97\approx 7$

Hence, a radioactive sample decay by 50% of it's initial value in 7 seconds

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Radioactive decay of sample

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