Bainbridge mass spectrograph and its explanation

Bainbridge mass spectrograph and its explanation

Bainbridge mass spectrograph

Bainbridge mass spectrograph consists of a discharge tube filled with the gas. Whose mass of ions is to be obtained. The beam of ions produced by the discharge tube passes through two fine parallel slits S₁ and S₂. And enters a velocity selector. The velocity selector consists of two parallel plates P₁ and P₂ between which a uniform electric field perpendicular to the direction of the ion beam, is produced by applying the potential difference between them. A uniform magnetic field is also produced in the same region perpendicular to both the ion beam and electric field.


Consider an ion having mass M, charge ne and moving with a velocity v. Let the electric field E and magnetic field B₁ are applied in a velocity filter. By adjusting the direction of the magnetic field, it is possible to make the electric force neE and magnetic force nevB₁ on the ions to be opposite to each other. If ions experience the equal and opposite force due to electric and magnetic fields such that neE = nevB₁, the ions beam will travel undeviated and pass through the slit S₃. Which is parallel and collinear with the slits S₁ and S₂. In such a situation, the velocity of the ion beam can be determined from the following equation.


neE = nevB₁

Or       v = E/B₁          …(1)

Brainbridge Mass Spectrograph

After emerging from slit S₃ the ions enter a semicircular chamber D. A uniform magnetic field B₂ is applied in this chamber, perpendicular to the plane of the incoming ion beam. An ion of charge ne experiences a magnetic force nevB₂ perpendicular to its path. An ion of charge ne experiences a magnetic force nevB₂ perpendicular to its path. As a result, the path of the ion would bend into an arc of a circle. If the mass of the ion is M, the centripetal force Mv²/r will be required to keep it along with a circular path of radius r. This centripetal force is supplied by the magnetic force.

Mv²/r = nevB₂

Or       r= Mv/neB₂      ..(2)

Substituting for v from equation (1) in it,

r = ME/ neB₁B₂         …(3)

Thus for a given value of E, B₁, B₂ ions of different values of M/ne will have different values of r.

Mass spectrum

After traveling through semicircular paths, ions strike a photographic plate situated along the order end of the diameter of the circle. Each set of ions having the same value of M/ne produce a linear trace on the plate. Thus a series of lines are obtained on the photographic plate similar to an atomic line spectrum. This spectrum is called mass spectrum.

The distance of the line from S₃ is equal to 2r, where r is defined by equation (3). The equation (3) shows that for ions of the same n, the distance is proportional to the mass M of the ion. Thus the scale of the spectrum is linear. By measuring these distances the isotopic masses of the ions can be determined. The densities of ions on the lines of the photographic plate determine the relative abundance of the isotopes. But the determination of relative abundance is not so accurate.

Example :

In a velocity filter, a magnetic field of .01 T is applied. In order to keep the path of the ions straight in the filter, a potential difference of 600 volts is applied between the plates is 0.05 m, then find the velocity of the ions emerging from the filter.

Solution : velocity of the ions passing through the filter

v= E/B

If the potential difference between the plates is V and the distance between them is d, then the electric field

∴  v = V/dB

Given V = 600 volt ; d = 0.05 m ; B = 0.01 T

v = (600/0.05 × 0.01 ) = 1.2 × 10⁶ m/s.


In this particular article Introduction of Brainbridge Mass Spectroscopy, we have studied Brainbridge Mass Spectrograph in detail.

Something Wrong Please Contact to Davsy Admin


Leave a Reply

Your email address will not be published.