= Solution
From <3/b>, the excited-state <probability> is
$$
\boxed{P_e(t)=\cos^2(gt).}
$$
For a single short <quantum measurement>[measurement] interval $\Delta t$ beginning in $|e,0\rangle$, requiring $P_e>0.9$ before the first departure from unity gives
$$
\boxed{\Delta t<\frac{\arccos\sqrt{0.9}}g
=\frac{0.32175\ldots}{g}.}
$$
A zero-photon outcome resets the conditional state to $|e,0\rangle$. After $n=T/\Delta t$ such tests, the <probability> of the all-zero-photon record is
$$
P_{\rm surv}(T)=[\cos^2(g\Delta t)]^n
\simeq e^{-g^2T\Delta t}.
$$
Thus a 90 percent survival requirement over a specified total duration $T$ is stronger than the single-interval condition: approximately $\Delta t<-\ln(0.9)/(g^2T)$ in the short-interval limit. The limiting preservation as $\Delta t\to0$ is the <quantum Zeno effect>.
The <quantum measurement>[measurement] instrument matters. If <photons> are absorbed and their first detection removes the excitation, this survival <probability> also equals the unconditional excited population. For ideal nondestructive <photon number>[photon-number] <quantum measurements>[measurements], excitation can transfer back after a one-photon result; the unconditional result after $n$ equal intervals is instead $P_e^{(n)}=[1+\cos^n(2g\Delta t)]/2$. It also tends to one for fixed $T$ as the intervals shrink. Neither interpretation guarantees indefinite 90 percent survival merely by imposing a 90 percent bound on each separate interval.
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