Chapter 20, Problem 63P

Solution

To calculate:The radius of curvature must a spectrometer have.

  $r=2500.95\times {10}^{-3}=2.5$ m

Given information:

From a mass spectrometer, it is difficult to separate molecules with nearly equal mass such as CO(28.0106u) and N₂(28.0134u). These two molecules are to be separated on the film by 0.50mm.

Formula used:

  $\frac{m{v}^2}{r}=qvB$ here r is radius and m is mass

Calculation:

The particle in mass spectrometer follow semicircular path.

From $\frac{m{v}^2}{r}=qvB$

Here $\frac{m{v}^2}{r}$ is the centripetal force and $qvB$ is the Lorentz force . Here Lorentz fore is provided by the centripetal force.

Now,

  $\frac{m}{r}=$ Constant

  ${\left(\frac{m}{r}\right)}_{co}={\left(\frac{m}{r}\right)}_{N_2}$

The distance between the point of entering to the point of striking the film is 2r

The two molecules are separated on the film by 0.50 mm. Difference in radii of two molecules is0.25mm

  $\left(\frac{28.0106}{r}\right)=\left(\frac{28.0134}{r+0.25\times {10}^{-3}}\right)$

  $28.0106(r+0.25\times {10}^{-3})=28.0134r$

  $28.0106r-28.0134r=-28.0106\times 0.25\times {10}^{-3}$

  $-0.0028r=-7.00265\times {10}^{-3}$

  $r=\frac{7.00265\times {10}^{-3}}{0.0028}$

  $r=2500.95\times {10}^{-3}=2.5\text{m}$