Question

In one type of mass spectrometer, ions having the same velocity move through a uniform magnetic field. The spectrometer is being used to distinguish 12C and 14C ions. The 12C ions move in a circle of diameter 45.4 cm. Use these atomic mass values: 12C, 12.0 u; 14C, 14.0 u.

Answer 1

Answer:

 r = 0.5297 m

Explanation:

In this exercise we use Newton's second law where the force is magnetic

         F = ma

centripetal acceleration

          a = v² / r

          F = q v x B = q v B sin θ

where the angle between the velocity and the magnetic field is 90º, therefore the sin 90 = 1

we substitute

          q v B= m v² / r

          r = $\frac{m v^2 }{qv B}$

the mass of each isotope is

12C

          m12 = 6 m_proton + 6 m_neutrons

          m12 = (6 1,673 +6 1,675) 10⁻²⁷

          m12 = 20.088 10-27 kg

14C

          m14 = 6 m_proton + 8 m_neutron

          m14 = (6 1,673 + 8 1,675) 10-27

          m14 = 23,438 10⁻²⁷ kg

in the exercise they indicate that the velocity of the two particles is the same, therefore with the initial data we can calculate the parameters that do not change in the experiment.

           $\frac{v}{qB} = \frac{r}{m_{12}}$

           v / qB = 0.454 / 20.088 10⁻²⁷

           v / qb = 2.26 10²⁵

this quantity remains constant, let's use the other data to calculate the radius

          r = 23.438 10⁻²⁷   2.26 10²⁵

         r = 5.297 10⁻¹ m

         r = 0.5297 m