From Part b, an atom of ^{3}H is more massive than an atom of ^{3}He.

- mass
^{3}He: 2p, 2e, 1n

$\mathbf{mass}{}^{\mathbf{3}}\mathbf{He}\mathbf{=}\mathbf{2}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0073}\mathbf{}\mathbf{amu}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{000549}\mathbf{}\mathbf{amu}\mathbf{)}\mathbf{+}\mathbf{1}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0087}\mathbf{}\mathbf{amu}\mathbf{)}$

**mass 3He = 3.02439 amu**

- mass
^{3}H: 1p, 1e, 2n

$\mathbf{mass}{}^{\mathbf{3}}\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0073}\mathbf{}\mathbf{amu}\mathbf{)}\mathbf{+}\mathbf{1}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{000549}\mathbf{}\mathbf{amu}\mathbf{)}\mathbf{+}\mathbf{2}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0087}\mathbf{}\mathbf{amu}\mathbf{)}$**mass 3H = 3.0252 amu**

$\mathbf{\u2206}\mathbf{m}\mathbf{=}\mathbf{mass}\mathbf{}{}^{\mathbf{3}}\mathbf{H}\mathbf{-}{}^{\mathbf{3}}\mathbf{He}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0252}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{02439}\mathbf{)}\mathbf{amu}$

**= 8.02 x 10 ^{-4} amu**

Precision of mass spectrometer:

- 6.022 x 10
^{23}amu = 1 g

The natural abundance of ^{3} He is 0.000137 %.

Based on your answer for part (b), what would need to be the precision of a mass spectrometer that is able to differentiate between peaks that are due to ^{3}He^{+} and ^{3}H^{+} ?

a) 1 × 10^{− 21 }g

b) 1 × 10^{−21} kg

c) 1 × 10^{−27} kg

d) 1 × 10^{−24 }g

e) 1 × 10^{−27 }g

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