Hello, it says that we must derive m = qB^2r^2/2V ..(volts)
using formula for kinetic energy and E= kq/r^2
thanks in advance for the help
A mass spectometer accelerates ions through a voltage, V, before the ions enter a uniform magnetic field, B, that makes them turn. Develop the forumla m= qB^2r^2/2V for the mass of the ion, using kinetic and charge energy. the radius of the turn is r and the charge is q
K.E. = qV when ion enters magnetic field: $\displaystyle \frac{mv^2}{2} = qV$
$\displaystyle \Rightarrow mv^2 = 2 q V$ .... (1)
In magnetic field: $\displaystyle F = \frac{mv^2}{r} = q v B \Rightarrow mv = r q B$
$\displaystyle \Rightarrow v^2 = \frac{r^2 q^2 B^2}{m^2}$ .... (2)
Substitute (2) into (1): $\displaystyle \frac{r^2 q^2 B^2}{m} = 2 q V$
and I'm sure you can make m the subject and simplify.