Question

Calculate the time of arrival of singly charged cations with masses 14, 15, 28, 32, 550, and 551, when the ions are accelerated with 10,000 V and distance to the detector is 1m

Answer 1

The potential energy of a charged particle in an electric field is related to the charge of the particle and to the strength of the electric field:

E_p=qU   ----(1)

where E_p is potential energy, q is the charge of the particle, and U is the electric potential difference (also known as voltage).

When the charged particle is accelerated into time-of-flight tube by the voltage U, its potential energy is converted to kinetic energy.

The kinetic energy of any mass = mv²/2 ----(2)

where v is the velocity and m is the mass

In effect, the potential energy is converted to kinetic energy, meaning that equations (1) and (2) are equal.

qU= mv²/2   -----(3)

The velocity of the charged particle after acceleration will not change since it moves in a field-free time-of-flight tube. The velocity of the particle can be determined in a time-of-flight tube since the length of the path (d) of the flight of the ion is known and the time of the flight of the ion (t) can be measured using a   transient digitizer.

V = d / t ---(4)

Substitute (4) in (3)

qU= m(d/t)² / 2----(5)

rearranging for t,

we get

t= (d/sqrt(2U)) * sqrt(m/q) –(6)

The above relation is used to calculate the time of flight.

Given:

d=1m , U=10000 Volts , q= charge of single cation=charge of electron =1.602*10^-19 C

note : [The given masses are in Daltons hopefully    (since units are not given)

convertion   1 dalton = 1.6605*10^-27 kg

in equation (6) d is in mts, U is in volts , mass is in kg, q is in Columbs, t is in sec]

substituing the above values in the equation (6) we get

mass in daltons	time in sec
14	2.6933*10-6 sec
15	2.7879*10-6
28	3.809*10-6
32	4.07*10-6
550	1.688*10-5
551	1.689*10-5