= Solution
In the intended additive-deposition model, A originally contributes $0.35\times2.2=0.77\,\mathrm{nm\,s^{-1}}$. At equal incorporated contributions, B also supplies $0.77\,\mathrm{nm\,s^{-1}}$, giving \b[$1.54\,\mathrm{nm\,s^{-1}}$]. Equivalently, the new total incorporated flux is $0.35+0.35=0.70$ times the calibrated total.
There is a composition-convention qualification. If the percentages represent <mass> flux and the <mass density> stays fixed, the preceding thickness calculation follows directly. If they instead represent atomic fractions, a constant atomic number density would justify the same result, whereas constant <mass density> alone does not. With atomic <masses> $M_A,M_B$ and constant <mass density>, the ratio would be
$$
\frac{R_{\rm new}}{R_{\rm old}}=\frac{0.35(M_A+M_B)}{0.35M_A+0.65M_B}.
$$
The PDF does not specify atomic versus <mass> percentages or the two elements. Thus \b[1.54 nm/s is the intended common flux-to-thickness approximation], and the formula above states the additional dependence if atomic fraction and constant <mass density> are meant literally.
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