thickness calculation:

object weight in **μg/cm ^{2}**

object density in **g/cm ^{3}**

Result = m.

Result = cm.

Result = um.

note:

μm = 10^{-6} m

mm = 10^{-3} m

cm = 10^{-2} m

As an example, for the 260 μg/cm^{2} ^{18}O target formed by the electrolysis of water enriched 97% in ^{18}O on a 12.7 μm Tantalum backing. The common general compund is Ta_{2}O_{5} and its density is around 8.2 g/cm^{3}. Ta has atomic mass = 181, and then 260 μg/cm^{2} O will prodcue 260/(18*5) * (181*2 + 18*5) = 1306 μg/cm^{2} Ta_{2}O_{5}, and then the thickness is 0.000159 cm.

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The numbers of Ta$_{2}$O$_{5}$ in a cube of cm$^{3}$ = NA * density / {a.m.u(Ta)$\times$2 + a.m.u(O)$\times$5}

the unit of a.m.u is equal to [g/mol]

the density is in [g/cm$^{3}$]

NA = Avogadro number = 6.022 $\times$ 10$^{23}$

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Suppose we know a reaction has cross-section = cs mb (mini-barn). And then we can estimate the total reaction events from the information of target. 'cross-section' is in the unit of area, and its concept is when the incident particle coming into this effective area then the reaction will happen. For classical cases, if the target is a sphere with radius = $R$, then the cross-section = $ 4\pi R$.

set

cs = $0.1$ mb = $10^{-28}$ cm$^{2}$, for a reaction with $^{18}$O target

n = 5.48$\times$10$^{22}$ = number of $^{18}$O in a cube of cm$^{3}$.

t = 2.44$\times$10$^{-4}$ cm = target thickness

beam_rate = 10$^{7}$ ions/sec

Reaction rate = beam_rate$( 1-exp(-n\cdot cs \cdot t) )$, according to reference(10.4)

roughly speaking we will have 50 events per hour.