Question

A simple mass spectrometer may include an electron ionization (EI) source and magnetic sector mass analyzer. In this type of instrument, singly charged ions are produced and accelerated through the slit to the analyzer by applying high potentials to accelerator plates. If an ion with mass 400 amu and charge z = 1 is accelerated by a potential of 4000 V, what is its kinetic energy (in J)?

Answer 1

K.E = 6.4 × 10⁻¹⁶

Velocity = 4.39 × 10⁴ m/sec

Step-by-step explanation:

From the given information:

The average K.E = P.E (potential energy)

Thus, K.E = q × V

K.E = 1.6 × 10⁻¹⁹ × 4000 V

K.E = 6.4 × 10⁻¹⁶

However,

`K.E = (1)/(2)mv^2`

`6.4 * 10^(-16) = (1)/(2)(400 * 1.66 * 10^(-27) ) * v^2`

`(6.4 * 10^(-16) )/((1)/(2)(400 * 1.66 * 10^(-27) ))= v^2`

`v^2=1.92771084 * 10^9\\ \\ v=√( 1.92771084 * 10^9) \\ \\ v = 43905.7 \\ \\ \mathbf{v = 4.39 * 10^4 \ m/sec}`