The electric field between the plates of the velocity selector in a Bainbridge m

Question

The electric field between the plates of the velocity selector in a Bainbridge mass spectrometer is 1.27×10^5 , and the magnetic field in both regions is 0.580 T. A stream of singly charged selenium ions moves in a circular path with a radius of 31.0 cm in the magnetic field.

PartA:
Determine the mass of one selenium ion.
m= ?kg
PartB:
Determine the mass number of this selenium isotope. (The mass number is equal to the mass of the isotope in atomic mass units, rounded to the nearest integer. One atomic mass 1u=1.66*10^-27kg.)

Thank you in advance~

Explanation / Answer

we have particle charged 1.6e-19 C (single charged ion) 1. calculate the velocity of the ion In order to move on straight line at the region where E and B are present , the ions mash have velocity v=E/B, then v=E/B=1.27e5/0.580=2.2e5 m/sec 2. Radius of movimg in magnetic field R=mv/qB the, m=RqB/v=1.3e-25 kg 3. to find the mass number N=m/1u=1.3e-25/1.66e-27=79

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