A radioactive sample decays by 63- of its initial value in 10s- It would have decayed by 50- of its initial value in

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Toppr Admin · Accepted Answer

Correct option is A- -7s-A radioactive sample decays by -63- in -t-dfrac-1-lambda-Hence -lambda-1-10-N-N-0e-lambda t-dfrac-N-0-2-N-0e-0-1t-0-1t-ln2-t-6-93s-approx7s-

A radioactive sample decays by $$63$$% of its initial value in $$10s$$. It would have decayed by $$50$$% of its initial value in

Correct option is A. $$7s$$
A radioactive sample decays by $$63$$% in $$t=\dfrac{1}{\lambda}$$
Hence $$\lambda=1/10$$
$$N=N_0e^{-\lambda t}$$
$$\dfrac{N_0}2=N_0e^{-0.1t}$$
$$0.1t=ln2$$
$$t=6.93s\approx7s$$

A radioactive sample decays by $$63$$% of its initial value in $$10s$$. It would have decayed by $$50$$% of its initial value in

Correct option is A. $$7s$$A radioactive sample decays by $$63$$% in $$t=\dfrac{1}{\lambda}$$Hence $$\lambda=1/10$$$$N=N_0e^{-\lambda t}$$$$\dfrac{N_0}2=N_0e^{-0.1t}$$$$0.1t=ln2$$$$t=6.93s\approx7s$$

Correct option is A. $$7s$$
A radioactive sample decays by $$63$$% in $$t=\dfrac{1}{\lambda}$$
Hence $$\lambda=1/10$$
$$N=N_0e^{-\lambda t}$$
$$\dfrac{N_0}2=N_0e^{-0.1t}$$
$$0.1t=ln2$$
$$t=6.93s\approx7s$$