Radioactive Decay Law

$N_x=N_0e^{-\lambda_x t}$, $N_y=N_0e^{-\lambda_y t}$. $\frac{N_x}{N_y}=e^{-(\lambda_x-\lambda_y)t}=\frac{1}{e} \Rightarrow (\lambda_x-\lambda_y)t=1 \Rightarrow t=\frac{1}{10\lambda-\lambda}=\frac{1}{9\lambda}$.

$\frac{A}{A_0} = \left(\frac{1}{2}\right)^{t/T_{1/2}} = \left(\frac{1}{2}\right)^{150/100} = \left(\frac{1}{2}\right)^{3/2}$ $= \frac{1}{2\sqrt{2}}$

As framed ($N_B/N_A = 1/e$), no option is correct, since B decays more slowly and $N_B/N_A$ only increases. If we take $\frac{N_A}{N_B} = \frac{1}{e}$, then $\frac{N_A}{N_B} = \frac{e^{-8\lambda t}}{e^{-\lambda t}} = e^{-7\lambda t}$ $\frac{1}{e} = e^{-7\lambda t} \Rightarrow -1 = -7\lambda t$ $t = \frac{1}{7\lambda}$ Note: the source states that no option is correct as the question is worded; the key accepts $\frac{1}{7\lambda}$ by taking the ratio $N_A/N_B = 1/e$.

Decay law: $N = N_0e^{-\lambda t}$
At $t = T_{1/2}$, $N = \dfrac{N_0}{2}$: $\dfrac{1}{2} = e^{-\lambda T_{1/2}}$
$\Rightarrow \lambda T_{1/2} = \ln 2$
$T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}$
($\dfrac{1}{\lambda}$ is the mean life.)