Mass spectrometer radius of the path

[1timeuse]

Suppose that an ion source in a mass spectrometer produces doubly ionized gold ions (Au2+), each with a mass of 3.27 x 10^-25 kg. The ions are accelerated from rest through a potential difference of 1.40 kV. Then, a 0.600 T magnetic field causes the ions to follow a circular path. Determine the radius of the path.

i tried to use

$$r = \frac {\sqrt{2mV}}{2q B^2}$$

i doubled the q because the question says the ions are doubly ionized. is q = 1.6 e -19 still or is it different with gold ions? my answer was 7.95 e -3 but I'm wrong according to my webassign.

 

[Tide]

Suppose that an ion source in a mass spectrometer produces doubly ionized gold ions (Au2+), each with a mass of 3.27 x 10^-25 kg. The ions are accelerated from rest through a potential difference of 1.40 kV. Then, a 0.600 T magnetic field causes the ions to follow a circular path. Determine the radius of the path.

i tried to use

$$r = \frac {\sqrt{2mV}}{2q B^2}$$

i doubled the q because the question says the ions are doubly ionized. is q = 1.6 e -19 still or is it different with gold ions? my answer was 7.95 e -3 but I'm wrong according to my webassign.

Your denominator also belongs under the radical.

 

[wais]

The formula you used, r = \frac {\sqrt{2mV}}{2q B^2}, is the correct one to use for finding the radius of the path of the ions in a mass spectrometer. However, the value of q will change depending on the charge of the ion. In this case, since the gold ions are doubly ionized, the charge would be 2 times the fundamental charge, so q = 2(1.6 x 10^-19) = 3.2 x 10^-19 C.

Plugging in the given values, we get:

r = \frac {\sqrt{2(3.27 x 10^-25)(1.40 x 10^3)}}{2(3.2 x 10^-19)(0.600)^2} = 0.017 m or 1.7 cm

So the radius of the path of the gold ions in this mass spectrometer would be approximately 1.7 cm. Make sure to double check your calculations and units to ensure accuracy.