Singly ionized (q=1.6X10-19 C) nuclei are shot through a velocity selector as pa

Question

Singly ionized (q=1.6X10-19 C) nuclei are shot through a velocity selector as part of a mass spectrometer. The electric field is 1.0x109 v/m and the velocity with which they exit the velocity selector is 7.3X101 m/s. 3) What magnetic field must exist inside the velocity selector? (9 pts) Assuming the magnetic field outside the velocity selector is the same as inside and the nuclei hit the wall of the spectrometer a distance y=0.145 m from the end of the velocity selector, what is the mass of the particles in kg? (8 pts) a. b. c. If singly ionized lead nuclei, mass 3.44X1025 kg, were shot through the system instead, at what distance from the end of the velocity selector will they hit the wall of the mass spectrometer? (8 pts) 2r1

Explanation / Answer

In the velocity selector

magnetic force = electric force

Fb = FE

qvB = E*q

B = E/v

magnetic field B = 10^5/(7.3*10^4) = 1.37 T

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(b)

In the spectrometer

magnetic force provides the necessary centripetal force

Fb = Fc

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

r = m*v/(q*B)

part b

mass m = r*q*B/v

radius r = y/2 = 0.145/2 = 0.0725 m

q = 1.6*10^-19

B = 1.37 T

v = 7.3*10^4 m/s

mass m = 0.0725*1.6*10^-19*1.37/(7.3*10^4)

mass m = 2.18*10^-25 kg

==========================

part c

distacne y = 2*r2 = 2*m*v/(q*B)

given mass m = 3.44*10^-25 kg

v = 7.3*10^4 m/s

q = 1.6*10^-19 C

distance y = 2*3.44*10^-25*7.3*10^4/(1.6*10^-19*1.37)

distance y = 0.23 m <<<<------ANSWER

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