3. Two singly-ionized atoms are analysed using a mass spectrometer. Initially, t

Question

3. Two singly-ionized atoms are analysed using a mass spectrometer. Initially, they pass through a velocity selector which has magnetic field strength 0.58 T and electric field strength 2.48 x 104 V/m. They then enter a detector region which has a uniform magnetic field with strength 0.2 T, perpendicular to the atoms' velocity a. What is the velocity of the atoms as they exit the velocity selector and enter b. One of the atoms collides with the detector at distance of 5.32 cm from c. The second atom collides with the detector at a slightly greater distance the detector region? where it enters. What is the mass of this atom? What can you conclude?

Explanation / Answer

First apply force balance on atoms in velocity selector

Fm = Fe

Fm = magnetic force = q*V*B

Fe = electric force = q*E

q*V*B = q*E

E = V*B

V = E/B = 2.48*10^4/(0.58)

V = speed of atoms = 42758.62 m/sec

Now when particle enters in detection region

again using force balance

Fm = Fc

q*V*B = m*V^2/r

m = q*B*r/V

Given that r = (5.32/2) cm = 2.66 cm

q = single ionized = 1 e = 1.6*10^-19 C

Using known values:

m = 1.6*10^-19*0.2*2.66*10^-2/42758.62

m = 1.99*10^-26 kg = mass of 1st atom

Part C.

Now if 2nd atom collides at a greater distance, then it means

r2 > r1

from above equation

m = q*B*r/V

we can see that mass is directly proportional to radius

So if atom striker at greater distance, than we can say that 2nd atom will be heavier than 1st atom.

Please Upvote. Let me know if you have any doubt.

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