Question

Assume our velocity selector is a bit leaky and lets the other 2 isotopes through to the mass spectrometer. 3. What is the spacing between the isotopes when they strike the detectors?

Just 3 and 4

An Iron-Nickel sample from a meteorite was placed into a velocity selector and mass spectrometer. Experimenters expect 3 primary isotopes: 56Fe, 58Ni, & 60Ni. Positive ions are accelerated by an applied potential (AV) of 3300 Volts which is the same potential difference used to create the electric field in the velocity selector, where d is 1.1 cm. 1. On the figure provided, in the velocity selector region, (A) draw the direction of the electric field. (B) Draw the necessary corresponding magnetic field to produce a velocity selector in the velocity selector region and, assuming the magnetic field is the same in the mass spectrometer, the mass spectrometer region. (C) Draw the approximate path for a positive VELOLITY ions within the mass spectrometer region. SELECTOR 2. Assuming the velocity selector is designed for singly- charged 58Ni, what must be the average magnetic field in this region? ov DETECTOARD FOR 1995 blece OME TEE Assume our velocity selector is a bit leaky and lets the other 2 isotopes through to the mass spectrometer. 3. What is the spacing between the isotopes when they strike the detectors? 4. Do we need to worry about doubly ionized isotopes entering the mass spectrometer? Defend your answer using math.

Answer 1

Electric Field Lines -3300 V OV + + + + + + - + + + ov Magnetic Field lon Detector N Circular Path of ions A y х

Part-(1)

(A) Direction of electric field in velocity selector is considered as direction of +z-axis as shown in figure (orange arrow lines )

(B) Positive ions deflected upwards in the assumed electric field .

If velocity of +ve ions is in the +y axis direction , in order to get downward deflection to balance the deflection due to electric field, magnetic field has to be applied in +x axis direction as shown in figure.

Magnetic field direction is out of page,shown as $\odot$ in figure

(C) Circular path of +ve ions in mass spectrometer region is shown as blue line

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Part-(2)

Magnetic field induction B applied to velocity selector is calculated from following equation

E = v B

where E is electric field intensity and v is speed of ions

Since ions are accelerated by 3300 V , speed of ions is calculated as follows

eV = (1/2) m v²

where e is charge on singly ionised ⁵⁸Ni ^, V is accelerating potential and m is amss of ion

$v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2 \times 1.602 \times 10^{-19}\times 3300}{58\times 1.66\times 10^{-27}}} \approx 10^{5} m/s$

Hence magentic field induction B = E / v = 3300 / ( 1.1 $\times$ 10^-2$\times$ 10⁵ ) = 3 T

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Part-(3)

Radius of singly ionised ⁵⁸Ni ion is claculated by equating magnetic force and centripetal force as follows

q v B = (m v² ) / R   

where R is radius of circular path .

We get radius R = ( m v ) / ( q B )   ..................(1)

R = ( 58 $\times$ 1.66 $\times$ 10^-27$\times$ 10⁵ ) / ( 1.602 $\times$ 10^-19$\times$ 3 ) = 20 $\times$ 10^-3 m

Radius R = 20 mm

We see from eqn.(1), Radius directly proportional to mass ,

Radius of ⁵⁶Fe= 20 $\times$ (56/58) = 19.34 mm

Radius of ⁶⁰Ni= 20 $\times$ (60/58) = 20.69 mm

Hence spacing between ⁵⁸Ni and ⁵⁶Fe = 2 $\times$ ( 20 - 19.34) mm = 1.32 mm

Spacing between ⁵⁸Ni and ⁶⁰Ni = 2 $\times$ ( 20.69 - 20) mm = 1.38 mm

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If we see from eqn.(1), Radius inversely proportional to charge q and directly proportional to mass.

Hence for doubly ionised ion, radius will be halved.

So we may not be knowing if the detected ion is doubly ionised or having half of mass.