The mass spectrometer is used on carbon isotopes for carbon dating. If the ^12 C

Question

The mass spectrometer is used on carbon isotopes for carbon dating. If the ^12 C isotope's radius of curvature is 25 cm, and the carbon isotope is singly ionized (q =), what is the required magnetic field B_2? a. 1.5 T b. 2.5 T c. 14 T d. 25 T e. 120 T Find the separation difference of ^12 C and ^14 C after being bent the half-circle as shown in the figure. (1 atomic mass unit = 1.66 times 10^-27 kg) a. 2.3 cm b. 1.3 cm c. 2.2 cm d. 4.2 cm e. 6.0 cm What magnetic field, B_2, would be required for a separation distance of 1.0 cm for ^U and ^U? Assume the same velocity selector speed of 800 km/sec. Is this a realistic setup?

Explanation / Answer

Q15. for mass spectrometer,

mv^2/r=q*v*B

==>B=m*v/(q*r)

where B=magnetic field

m=mass of the ion=12.011 amu=1.9938*10^(-26) kg

v=speed of the velocity selector=800 km/sec=8*10^5 m/s

q=charge of the ion=1.6*10^(-19) C

r=radius of curvature=0.04 m

then B=1.9938*10^(-26)*8*10^5/(1.6*10^(-19)*0.04)=2.4922 T

so option B is correct.

Q16.

radius of curvature=m*v/(q*B)

distance after being bent the half circle =difference between diamter of curvature

=(2*v/(q*B))*(m2-m1)

where m2=mass of 14C

m1=mass of 12C

so distance=(2*8*10^5/(1.6*10^(-19)*2.5))*(14-12)*1.66*10^(-27)=0.01328 m

=1.328 cm

hence option B is correct.

bonus:

separation distance=(2*v/(q*B))*(m2-m1)

==>(2*8*10^5/(1.6*10^(-19)*B))*(238-235)*1.66*10^(-27)=0.01

==>0.0498/B=0.01

==>B=0.2008 T

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