In: Physics
Suppose the electric field between the electric plates in the mass spectrometer of Figure 27-33 in the textbook is 2.64×104 V/m and the magnetic fields are B=B′=0.36T. The source contains carbon isotopes of mass numbers 12, 13, and 14 from a long dead piece of a tree. (To estimate atomic masses, multiply by 1.66×10−27kg.)
How far apart are the lines formed by the singly charged ions of each type on the photographic film?
What if the ions were doubly charged?
by force balance on ion in velocity selector
Fe = Fb
q*E = q*v*B'
v = E/B'
Now when ions passes through magnetic field on to photographic film, radius of their paths will be
r = m*v/(q*B)
Diameter will be
d = 2*r
d = 2*m*v/(q*B)
So, d = 2*m*E/(q*B*B')
given , E = 2.64*10^4 V/m
B = B' = 0.36 T
q = single charged = 1.6*10^-19 C
here, m12 = mass of carbon 12 = 12*1.66*10^-27 kg
m13 = mass of carbon 13 = 13*1.66*10^-27 kg
m14 = mass of carbon 14 = 14*1.66*10^-27 kg
So,
d12-13 = d13 - d12 = 2*m13*E/(q*B*B') - 2*m12*E/(q*B*B')
d12-13 = [2*E/(q*B*B')]*(m13 - m12)
Using known values:
d12-13 = [2*2.64*10^4/(1.6*10^-19*0.36*0.36)]*(13*1.66*10^-27 - 12*1.66*10^-27)
d12-13 = [2*2.64*10^4*(1.66*10^-27)/(1.6*10^-19*0.36*0.36)]
d12-13 = 0.004227
In two significant figures
d12-13 = 0.0042 m
Now,
d13-14 = d13 - d12 = 2*m14*E/(q*B*B') - 2*m13*E/(q*B*B')
d13-14 = [2*E/(q*B*B')]*(m14 - m13)
Using known values:
d13-14 = [2*2.64*10^4/(1.6*10^-19*0.36*0.36)]*(14*1.66*10^-27 - 13*1.66*10^-27)
d13-14 = [2*2.64*10^4*(1.66*10^-27)/(1.6*10^-19*0.36*0.36)]
d13-14 = 0.004227
In two significant figures
d13-14 = 0.0042 m
Part B
when ions are doubly charged, then
q = 2*1.6*10^-19
then
d12-13 = [2*2.64*10^4*(1.66*10^-27)/(2*1.6*10^-19*0.36*0.36)]
d12-13 = 0.002113
In two significant figures
d12-13 = 0.0021 m
And
d13-14 = [2*2.64*10^4*(1.66*10^-27)/(2*1.6*10^-19*0.36*0.36)]
d13-14 = 0.002113
In two significant figures
d13-14 = 0.0021 m
So, final answers are
Part A: d12-13, d13-14 = 0.0042, 0.0042 m
Part A: d12-13, d13-14 = 0.0021, 0.0021 m
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