Question

An acceleration voltage of 30.00 kV is used to accelerate ions into the 1.657 m flight tube of a time-of-flight mass spectrometer. If two singly, positively-charged ions, one with a m/z of 5068.664 and one with a m/z of 5077.857, are accelerated at the same time into the flight tube, what will be the difference in the arrival time of the two ions at the detector? Enter the difference in arrival time as a positive value.

Answer in ns.

Answer 1

as these are singly charged ions,

hence, charges of ions ( q₁ =q₂ ) = 1.6*10^-19 C

m/z (mass to charge ratio) of 1st ion = 5068.664

hence, mass of 1st ion (m₁ ) = 5068.664*mass of single charge = 5068.664 * 1.66*10^-27 Kg =8.41398224*10^-24 Kg

where, mass of single charge = 1.66*10^-27 Kg

potential energy of 1st ion = q₁ *V = 1.6*10^-19 *30*10³ J = 4.8*10^-15 J

kinetic energy by this potential = 1/2*m₁ *v₁² = 4.8*10^-15 J

hence,

0.5*8.41*10^-24 *v₁² = 4.8*10^-15

v₁² = 1.140957959*10⁹

v₁ = 3.37780692*10⁴ m/s

time taken by 1st ion = length of tube / velocity of 1st ion = 1.657/ 3.37*10⁴ s

t₁ = 4.90554978*10^-5 s -----(1)

similarly, for 2nd ion

m/z (mass to charge ratio) of 2nd ion = 5077.857

hence, mass of 2nd ion (m₂ ) = 5077.857* mass of single charged = 5077.857*1.66*10^-27 Kg =8.42924262*10^-24 Kg

where, mass of single charged = 1.66*10^-27 Kg

potential energy of 2nd ion = q₂*V = 1.6*10^-19 *30*10³ J = 4.8*10^-15 J

kinetic energy by this potential = 1/2*m₂ *v₂² = 4.8*10^-15 J

hence,

0.5*8.43*10^-24 *v₂² = 4.8*10^-15

v₂² = 1.13889*10⁹

v₂ = 3.374747*10⁴ m/s

time taken by 2nd ion = length of tube / velocity of 2nd ion = 1.657/ 3.374747*10⁴ s

t₂ = 4.909996 *10^-5 s -----(2)

from 1 and 2

difference in arrival time = t₂ -t₁ =  4.909996 *10^-5 s - 4.90554978*10^-5 s = 4.4102*10^-8 s = 44.102 ns