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(C) The law of radioactive decay is given by $N = N_0 (1/2)^{t/T_{1/2}}$,where $T_{1/2}$ is the half-life.Given that the undecayed fraction $N/N_0 = 1

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$31/32$

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(C) The law of radioactive decay is given by $N = N_0 (1/2)^{t/T_{1/2}}$,where $T_{1/2}$ is the half-life.Given that the undecayed fraction $N/N_0 = 1 - 7/8 = 1/8$ is incorrect based on the phrasing 'undecayed fraction is $7/8$'. Let $N/N_0 = 7/8$ at $t = 6$ days.Using $N = N_0 e^{-\lambda t}$,we have $7/8 = e^{-6\lambda}$.Taking the natural logarithm on both sides: $\ln(7/8) = -6\lambda$,so $\lambda = -\frac{1}{6} \ln(7/8) = \frac{1}{6} \ln(8/7)$.Now,for $t = 10$ days,the undecayed fraction is $N/N_0 = e^{-10\lambda}$.Substituting $\lambda$: $N/N_0 = e^{-10 \cdot \frac{1}{6} \ln(8/7)} = e^{-\frac{5}{3} \ln(8/7)} = (8/7)^{-5/3} = (7/8)^{5/3}$.Calculating this value: $(7/8)^{1.666} \approx 0.875^{1.666} \approx 0.798$.However,if the question implies $N/N_0 = 1/8$ (decayed fraction $7/8$),then $1/8 = (1/2)^{6/T_{1/2}} \Rightarrow (1/2)^3 = (1/2)^{6/T_{1/2}} \Rightarrow T_{1/2} = 2$ days.For $t = 10$ days,$N/N_0 = (1/2)^{10/2} = (1/2)^5 = 1/32$.The undecayed fraction is $1/32$. If the question asks for the decayed fraction,it is $1 - 1/32 = 31/32$.

If the undecayed fraction of a radioactive sample is $7/8$ after $6$ days,what will be the undecayed fraction after $10$ days?

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